Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: F4UDOA on January 03, 2002, 11:03:00 AM
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Heya's,
I posted these docs on my Web page a while ago. However I didn't break out these two pages of comparison charts to be viewed. So here they are.
The first one shows comparisons of Drag Coefficients, Wing Loading, Power Loading etc.
(http://members.home.net/markw4/alliedchrts2.jpg)
Second is a Performance comparison.
(http://members.home.net/markw4/alliedchrts1.jpg)
What stands out in these charts to me are these things.
1. The P-38J has two 1600HP engines and has a VMAX of only 415MPH.
2. The Performance of the P-47D. It has outstanding Speed from sea level up to 30K at 440MPH.
3. The Drag Coefficients of these A/C. Based on these and available power I find it hard to believe that F4U does not accelerate better based on available power versus low drag. Could someone expain this?
4. Are AH stall speeds too high accross the board?
5. Based on the Navy's Range calculation the F4U-4 has a longer range than the F4U-1. In AH it has a shorter range. Why?
6. The P-38J has a relatively short range in comparison to other A/C. Why? I thought the P-38 had longer legs. Looking at the P-38 Manual it would seem that it consumed a great deal of fuel compared to the F4U and only carried marginally more fuel.
Any comments?
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F4UDOA, I am SO glad you reposted this info. I searched for it for hours last week, never found it.
FWIW, I remember someone pointing out the P47D as having the "wrong" engine, which they said accounts for the outstanding speed it gave. BS in my opinion, but that is what they said.
One thing I wonder when look at the charts is if the ones who did the testing on the birds had an F6F with the "revised" airspeed indicator? Grumman chief test pilot tested F6F and Corsair, found that even in formation flying, the Corsair "indicated" about 20 knots faster than the F6F, so they worked on the pitot tube and such til the airspeeds were identical when the planes were in formation. He also said that in testing side by side, the planes had identical performance and speeds at the low and high blower stages from 5K feet up to service ceiling, with the difference in main stage blower being the way the air was routed to the carb in each plane.
The tests were conducted in the summer of 1943, with the planes being an F6F-3 and an F4U-1D serial #17781.
From what I gather reading his summary, the F6F was a 400mph plane just like the F4U, and not 20-odd mph slower like all the tests show, and the sims model.
Just my thoughts and 2 cents worth on the subject.
Sorry I strayed so far off topic......<S>!
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Well... first of all why did they use a P-38 J version. This kind of points at "old" data.
I think the 38's had some serious fuel/range problems to begin with. This is why Lindenburg was sent to show them how to properly trim and use engine management to get much more range out of their planes.
Other than that I quess one of the experten we see such as Widewing and co can comment w/much more Technical info I'm sure. :)
xBAT
P.S. F4UDO..what is your web page addy? The link on your profile doesnt seem to work.
[ 01-03-2002: Message edited by: batdog ]
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Neat stuff.
From the weight I would guess that p47D is a razorback. The notes imply it only has 1 dt instead of 3 so I guess its without the wing pylons too. Baugher lists the R2800-63 as 2300hp, not the 2600 here. Maybe this is one of them hot-rod jugs :). Anyway that performance is pretty impressive, even with overload ammo.
Range seems pretty wierd. Unless there is a digit missing it says the P51b has less range than P47!?
<edit> Just noticed internal fuel for the P47 was 305 gallons, which I thought means bubble-top too. So i dunno whats up with that.
Wow, take off run in a Jug is twice as long as f4u/f6f. Is that from flaps, or a big difference takeoff speeds?
[ 01-03-2002: Message edited by: Regurge ]
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Originally posted by batdog:
[QB]Well... first of all why did they use a P-38 J version. This kind of points at "old" data.
I think the 38's had some serious fuel/range problems to begin with. This is why Lindenburg was sent to show them how to properly trim and use engine management to get much more range out of their planes.
Other than that I quess one of the experten we see such as Widewing and co can comment w/much more Technical info I'm sure. :)
QB]
Well, I'm at the office today, so my reference material is not with me. However, I should mention that Lindbergh did not introduce a cruise method that was not already in the P-38 Pilot's Manual! He merely used settings that were already defined, while the rest of the 475th FG preferred to cruise in Auto-rich, at high RPM. This allowed them to respond faster to enemy aircraft, but used up to 30% more fuel.
It should be noted that when using max range power settings in the P-38 (J or L), your airspeed was bog slow. Yeah, it had comparible range to the Mustang, but you arrived later. Another issue was the need to step up speed and get those props into fine pitch BEFORE you found yourself in a fight. In the ETO, this typically happened once the fighters crossed into known enemy operating areas, or about 2/3rds of the way to the target. OTOH, the Mustangs cruised at higher speeds and were always better prepared (in terms of aircraft power and prop settings) for sudden combat. Having to power-up well in advance of combat diminished the Lightnings range by at least 10%. Things improved for the P-38 when the L arrived, because of "single lever" powerplant controls, which made getting the props and engines pushed up a far less complex and time consuming procedure (when seconds count, no one wants to be fiddling around with seperate prop and throttle controls). Still, the extended range cruising speed remained well below that of the P-51. As you all know, it is vitally important to enter a potential combat area with plenty of airspeed. The Mustangs could and did, the Lightnings could, but wouldn't if they wanted to have enough gas to get home. All of this factored into the general data used by the 8th AF to decide on which fighter would become the principle long-range escort.
Another fact: Lindbergh spent time with P-47 groups and Marine Corsair squadrons teaching them how to extend the range of their fighters as well. Here again, the pilots tended towards high speed cruising in Auto-rich.
My regards,
Widewing
[ 01-03-2002: Message edited by: Widewing ]
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I thought the P-47D-30 and 40 which had the 2,595/2,600 hp rating with water injection had 370 gallons of internal fuel? were the planes not filled to full internal capacity? is there more data on the a/c that goes along with this chart? thanks F4UDOA
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Oops bolilo just reminded me the incresed fuel load in bubble-tops was 370, not 305 gallons. So i guess it is a razorback.
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Hi F4UDOA,
>1. The P-38J has two 1600HP engines and has a VMAX of only 415MPH.
The P-38J also has the highest drag coefficient, which counters its high power. Here's the list sorted for cubicroot(P)/Cd, which gives an impression of top speed relations (at sea level, since power is given at sea level):
F7F
F2G-1
XF8F
F4U-4
P-51B
F4U-1
P-47D
F6F-5
P-38J
(F6F-5 and P-38J are sperated by a big gap from the rest.)
>2. The Performance of the P-47D. It has outstanding Speed from sea level up to 30K at 440MPH.
It's interesting to compare the effect of the P-47D's turbosupercharger to that of the engine-driven supercharger of the P-51B: The P-51B is faster at low altitude and around each supercharger stage's critical altitude, but slower in the middle between the critical altitudes. The P-47D has no such performance "valley".
>3. The Drag Coefficients of these A/C. Based on these and available power I find it hard to believe that F4U does not accelerate better based on available power versus low drag. Could someone expain this?
Drag determines top speed. Power to weight ratio determines acceleration at zero airspeed. At low airspeeds, power to weight ratio dominates, at high airspeeds, drag dominates. Here's the list sorted by power (at sea level) to (gross) weight ratio:
XF8F
F2G-1
F7F
F4U-4
P-38J
F4U-1
P-47D
F6F-5
P-51B
Note that of the "2nd generation" fighters, the P-38J is at the top while the P-51B is at the bottom.
Regards,
Henning (HoHun)
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Just out of curiosity F4UDOA, when you refer to the range of the F4U-4 vs the F4U-1, are you relating both to their max range engine settings? I haven't tested it, just looking for some clarification is all. Thanks.
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HoHun,
I agree completely (Like I have a choice weather I can agree with the laws of Physics).
But having said that the brick wall of drag really hurts the P-38 even in the thin Air up high. You can tell based on it's climb at high alt. that it is certainly not lacking power. But as you said with high drag top speed is limited.
The Acceleration of the F4U really has me befuddled. I need to break out my Aerodymics book to get the equation but with a Power to weight slightly worse than the P-38J and a Cdo of .020 I would expect the acceleration to be very good. It's climb suffered because of some CLmax problems until a Paddle blade prop was adopted but the requirements for acceleration are fairly simple. Power to weight minus drag. Based on this I would expect the F4U-1 to be close to the P-38L in acceleration. I will throw some numbers around and post them.
Sundog,
No, I'm not basing it on Max rated settings. I'm looking at the bottom of the second chart posted where it shows based on the Navy's range calculator that the F4U-4 has a longer range than the F4U-1. The formula is shown at the bottom of the page.
Eddiak,
Do you have that test between the F4U and F6F done by Grumman? I have never seen a full copy. I do believe the F6F was a 400MPH fighter. I have my doubts about it being as fast as the F4U at 20K altought I know it close. In the test with the Zero and FW190 the F6F-3 and F6F-5 both reached 400MPH. The F6F-5 reached 409MPH at 20K with the F4U-1D hitting 413MPH. The F4U's big advantage is at sea level.
Bolillo and Batdog,
Here is the link to my homepage. The entire report is posted there. the is some more info on the P-38J also. Charts, HP, Manifold pressure etc. it is under the F4U comparative analysis link. It is 6 meg and in Adobe form. I would download it first if I were you :)
F4UDOA's web page (http://members.home.net/markw4/index.html)
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F4u-4 and F4U-1 ranges. If I read that chart correctly it says 237 gallons for the F4u-1. I believe the AH one is an earlier model with an additional fuel tank that carries 361 gallons internally.
Hooligan
[ 01-03-2002: Message edited by: Hooligan ]
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Pyro told me part of the F4U acceleration puzzle at the con.
Sometime during the -1d production run they switched to a fatter "paddle-type" propellor. The propellor on -1s and early -1Ds did not have very good efficiency in the climb speed range. The new propellor resulted in slightly less top-end speed but markedly better climb and low-speed acceleration.
So, to some extent earlier F4Us have disapointing acceleration and climb due to poor prop efficiency at lower speeds.
Hooligan
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Hi F4UDOA,
>Based on this I would expect the F4U-1 to be close to the P-38L in acceleration.
Acceleration depends on specific excess power, which is expressed as a climb rate.
Fortunately, for the aircraft in the F4U-4 comparison, we have specific excess power for 1 G flight at best climbing speed recorded in shape of the climb rate graphs :-)
Based on this, I can say that the P-38J will accelerate faster from its best climb speed than the F4U-1. (A slight inaccuracy may result from different speeds of best climb, but at sea level the P-38J has 3700 fpm specific excess power available to compared to just 3100 fpm of the F4U-1. The P-38 is clearly ahead there.)
Another specific excess power data point is provided by the top speed in 1 G flight - obviously, when no further acceleration is possible, the plane has no excess power available.
In other words, at 338 mph will F4U-1 will out-accelerate the P-38J since the latter won't accelerate at all :-)
Somewhere in between in the break-even point, where the lower drag of the F4U-1 will begin to pay off and overcome the power advantage that enabled the P-38J to accelerate faster initially.
Regards,
Henning (HoHun)
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All the numbers point to a need for a one plane Air Force....
The F7F :D
Bombs, torps, rockets, cannon, .50cals, radar for night work, speed and power :eek:
Yeh, I know "But it did not see combat". I still want it , designed, production line built (yep small #'s) in WWII, and not a jet.
Yes, with perk points.
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Hooligan,
I'm surprised Pyro said that. The prop change has been my personal battle cry of the F4U for almost a year. The new prop blade design albeit three bladed was the 6501A-0 as listed in the charts. It says right in the front of the F4U manual that the new prop design increases performance over previous models and should be used whenever possible. If you look at the F4U Vrs P-51 or FW190 test it mentions the prop change and the increase in performance there as well. It just took until Version 1.08 to make it into AH. However acceleration is another subject.
HoHun,
I have seen the charting method for excess power to climb. One line represents Horsepower available and the next is HP used to attain a certain speed. The gap between the two is climb. My issue is that climb has more variables than just Thrust-Drag/ Mass.
Climb has Lift coefficients. And the P-38 has a high Max Cl because of high aspect ratio where as the F4U has a relatively moderate Max Cl(no Flap with prop in case Niklas is reading) of 1.48. This reduces the climb of the F4U in relation the P-38. So the 3100FPM vrs the 3700FPM are really not directly proportion to acceleration. So if you calculate Thrust-Drag / Mass at a given speed say 150MPH I am will to bet that F4U does almost as well as the P-38 despite not being able to climb.
Any thoughts?
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Hiya F4UDOA,
I've submitted requests for copies of the F6F-3 tests conducted at the Naval Air Test Center in Patuxent. Maryland.
Whether or not I am able to get copies remains to be seen.
At the very least I am hoping they can redirect me to someplace that might still have copies.
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My issue is that climb has more variables than just Thrust-Drag/ Mass. Climb has Lift coefficients.
Lift coefficients only play a part in induced drag. Climb rate *is* (T-D)/W * V, whereas acceleration is (T-D)/Mass. Note that weight is related to mass by the gravity constant.
And the P-38 has a high Max Cl because of high aspect ratio where as the F4U has a relatively moderate Max Cl(no Flap with prop in case Niklas is reading) of 1.48. This reduces the climb of the F4U in relation the P-38. So the 3100FPM vrs the 3700FPM are really not directly proportion to acceleration.
You're right, but not because of lift coefficients, because climb speeds are different.
So if you calculate Thrust-Drag / Mass at a given speed say 150MPH I am will to bet that F4U does almost as well as the P-38 despite not being able to climb.
Acceleration is directly related to climb rate *at the same speed*. Let's say the F4u makes 3100 fpm (52 ft/s) @ 155 mph (227 ft/s) and the P-38 makes 3700 fpm (62 ft/s) @ 180 mph (264 ft/s).
To convert those climb rates to accelerations, you divide by V and multiply by gravity constant of 32.2 ft/s^2
P-38 @ 180 mph: a = 7.56 ft/s^2 (straight and level)
F4u @ 155 mph: a = 7.38 ft/s^2 (straight and level)
Whether the F4u could out-accelerate the P-38 at 155 mph depends on whether the P-38 can climb 3100 fpm at that speed.
[ 01-04-2002: Message edited by: wells ]
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thank you for the site and info F4UDOA, going to check it out. I am just curious, but why is it that everytime I download and then open a document it says it is "too large for note pad to open would you like word pad to read this for you", and then it is a bunch of characters that I do not understand? in reference to F4UDOA's document on "F4U comparative analysis" thanks
[ 01-04-2002: Message edited by: bolillo_loco ]
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I suspect that these test datas are from the planes used at the Fighter Conference?
I seem to remember having similar discussion with F4U about the P-47's HP.
The P&W R2800-63 was used on the D-11 to D-20 and was from factory set to max 2300HP.
However, it was common to increase the max MAP allowed, which is very likely what have happened to this one, so that it was producing 2600HP, while using WEP.
From P-47D-26 onwards, this was done at the factory as well.
Daff
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bolillo_loco
If I am correct your problem is that you are trying to read the document with the wrong program. For the .pdf file you need to download the free version of adobe acrobat. Otherwise for any other file type I would suggest Microsoft Word.
Hooligan
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yes thank you hooligan that worked for me :) I will be back to wine later now that I can read it :D
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I was gonna ask where Wells got his ROC equation, it looked a bit weird to me. Then I noticed that I was just confused with the imperial system (used to metric.) Duh.
Anyways, F4UDOA: do you have the frontal area numbers available that were used in calculating the drag coefficient? Since the drag is the product of the coefficient and the area, I guess the Cd and power alone would not tell the whole story about acceleration. Actually it would be real cool to see a comparison of those fighters based on the total drag and power, but I have never stumbled upon that kind of a table.
BTW, here is a nice link:
Beginner's guide to aerodynamics (http://www.grc.nasa.gov/WWW/K-12/airplane/bga.html)
Nice site, loads of good stuff about planes if you are like me and don't know the first thing about aerodynamics =)
Lapa
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Lapa, those drag coefficients are based on wing area.
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Hi F4UDOA,
>Climb has Lift coefficients. And the P-38 has a high Max Cl because of high aspect ratio where as the F4U has a relatively moderate Max Cl(no Flap with prop in case Niklas is reading) of 1.48. This reduces the climb of the F4U in relation the P-38.
Maxmimum lift coefficient in a climb would be reached at 1 G stall speed. That's much slower than best climb speed, and it's less efficient for climbing because the maximum lift is achieved at very high drag.
The best climb rate is reached at maximum excess power.
A first approach to find out what that means would have to look at the maximum ratio of lift to drag where the aircraft flies with least drag. This is reached at much smaller angles of attack than maximum lift.
However, since propeller efficiency increases with speed (in the low speed range at least), the best climb rate will be achieved at even higher speeds than that of best lift to drag, and accordingly at a yet smaller angle of attack.
So, the second approach will be to determine the point where the difference between usable power and speed-dependend drag is largest - that's the speed of best climb.
Sustained climb means flight at 1 G, so initial acceleration in level flight is indeed perfectly proportional to climb rate at the same speed :-)
The only way for a F4U-1 to accelerate away from a P-38J is to go into a steep dive to get the speed up into the region where total drag is more important than power-to-weight ratio.
Regards,
Henning (HoHun)
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the p38 accels when past 250mph? wow! :eek:
:D
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I compared this data to the AHT and some manual data I have. Generally the data about the Navy fighters (F4U-1, F4U-4 and F6F-5) appears to be bit more optimistic than the data in the AHT. There a little differences in the ratings; 59" vs 60" for the B series R-2800 and AHT's rating for the C series is probably at 64". But loadings are very similar (except some loadings for the F6F). Rammed critical altitudes are also a bit strange, they don't behave like they should.
The data for the Army fighters varies abit more. The AHT claims just around 13500lbs weight for the P-47D (up to D-25) at similar loading but a closer look to the AHT's data revealed that there is an error in the ammo loadings for the P-47D, ammunition should weight about 1057lbs not 664lbs. So weights for the P-47D are actually very similar in both sources. Generally rating for the R-2800-63 should be 2300hp@58" WET but 2600hp@64" WET rating certainly existed for the later B series P-47D engines and was used at least in the pacific. Again rammed critical altitudes are a bit strange.
The weight for the P-38J seems to be very low, with full ammo and 300 gallons of fuel it should weight atleast 16800lbs (actually a bit over 17000lbs according to the manual). The performance apppears to be it on low side if the weight is correct; at 16415lbs the P-38J should be able to climb very close to 4000fpm also speed is a bit low. The critical altitudes are again a bit strange but these can be partially explained by plane's behaviour to reach it's top level speed below critical altitude.
The engine model and rating information for the P-51B are very strange. The claimed MAP is for WER and critical altitudes looks like the values of the V-1650-3. But text claims that ratings are MIL and that engine is the V-1650-7 also hp values are closer or exactly for the V-1650-7 at MIL. Low altitude speed and climb can be be for both engines, the V-1650-7 actually did about same hp at MIL rating as the V-1650-3 at WER at low altitude. Also the P-51 typically reached it's top speed at 61" (above rammed critical altitude).
The Radius of action comparison is quite questionable, the P-38J and P-51B should have a better radius than the P-47D.
Overall it's a interesting set of data but it appears that it is not a real test report but a collection of data and sources are not claimed.
Then couple notes about level flight acceleration calculations. IMHO there is no quick and easy way to calculate correct accelerations; we should know propeller efficiency for given speed range and altitude, Cl values for given speed range and weight and how drag changes when Cl changes. Actually it is pretty damn difficult... Some of the data is available for the planes like the P-51 (total drag, drag/speed/Cl relations) but generally propeller part is a problem. Also exhaust thrust should be counted.
gripen
[ 01-04-2002: Message edited by: gripen ]
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All those chart and number things are neat, but this is still just a plain old whine. ;)
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Hi Gripen,
>IMHO there is no quick and easy way to calculate correct accelerations
Acceleration from low speed can be estimated from climb rate with good accuracy.
Here's an excess power chart for a generic WW2 propeller fighter:
(http://members.aol.com/hohunkhan/powerbalance.gif)
(Generic propeller efficiency curve, exhaust thrust figured in, ram effect neglected.)
The left hand limit is defined by the stall.
Where the total drag curve intersects the available power curve, the aircraft has reached its level top speed.
Where the difference between available power and total drag is the greatest (i. e. the excess power graph has its maximum), the best climb rate/acceleration is reached.
The interesting thing is that the excess power graph is rather shallow around the maximum, so that speed variations only have a minor impact on the climb rate/acceleration.
In other words, the relative acceleration of two WW2 propeller fighters at the same speed can be safely estimated from their relative climb rates even if these climb rates are achieved at different speeds.
Regards,
Henning (HoHun)
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Hohun,
Although prop efficiency increases with speed, thrust decreases, since
Thrust = Power * e / V
If you plot thrust instead of propeller efficiency, you will find that the greatest excess thrust is at the stall speed, where the best angle of climb is, not the best rate. As you noted, the drag curve is flat at that point, so increasing speed will result in a higher climb rate up to the point where excess thrust is decreasing faster than speed is increasing.
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HoHun,
English is not my native language so I might use wrong terms in following. As Wells noted we are interested about thrust/drag/weight(mass) relations not directly about power. Another problem is that climb rate/speed is a pretty static flying form but during acceleration relations between thrust and drag change when speed increases and in addition there are big differences between best climbing speeds like in the case of the P-38 (around 160mph IAS) and P-51 (around 200mph IAS).
It is pretty easy to create generalized models and I believe that at low speeds they give acceptable results but as real world test showed in the case of the P-38 and P-51, things are very different when speed increases also altitude can make a big difference.
gripen
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Hi Gripen,
>As Wells noted we are interested about thrust/drag/weight(mass) relations not directly about power.
The point I was trying to investigate is the relation between excess power of aircraft with a different speed of best climb, so the power balance chart was the way to go.
As Wells and you both pointed out, the shape of the thrust graph of course is different from that of the power graph. From each aircraft's best climbing speed, initial acceleration would be quite different, with the slower aircraft of course acclerating more quickly if their climb rates are similar.
However, what the power curve tells us is that when estimating acceleration at equal speed in the climb speed range from the climb rate, the exact speed of best climb will make little difference.
>there are big differences between best climbing speeds like in the case of the P-38 (around 160mph IAS) and P-51 (around 200mph IAS)
At sea level, that's 71.5 m/s true air speed for the P-38 and 89.3 m/s true air speed for the P-51. You can see from the above diagram (best climb speed 76 m/s) that the span between the two is easily on the plateau where excess power changes little.
Comparing sea level climb rates of 3730 fpm for the P-38J to the 3410 fpm for the P-51B, it safe to conclude that the P-38 accelerates better at the P-38's speed of best climb - judging from the above graph, the P-51 might lose about 5% of its climb speed by going slower, leaving the P-38 with a 15% advantage in climb/acceleration.
At the P-51's speed of best climb, the aircraft are matched closer - it's the P-38 that loses about 5% of its climb rate there, leaving it with a meagre 4% advantage in climb acceleration.
At even higher speeds, excess power drops faster, so I would expect the P-51's acceleration to match and finally exceed that of the P-38J's somewhere between 200 mph and 240 mph.
>also altitude can make a big difference
I think the biggest impact of altitude is by the different power characteristics of the aircraft engines. If you have climb rates at equal altitudes, you still can compare them with good confidence to estimate the relative acceleration (in the climb speed region).
Regards,
Henning (HoHun)
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HoHun,
Well, excess power is not what we are looking for because the propeller is the limiting factor despite what ever amount of power is available. Anyway, I admited allready that it is possible to get acceptable idea of relative acceleration of the planes from the climb rates at climb speeds and your conclusions are pretty well in line with the real world tests (except that you appear to use military rating values for the P-51 and WER values for the P-38). At higher altitude the P-51 reaches acceleration of the P-38 faster because it has relatively more thrust available at higher speeds (exhaust thrust and better propeller efficiency).
But earlier you wrote that:
>Acceleration from low speed can be estimated from climb rate with good accuracy
So how can you estimate accurate acceleration (value of m/s2) at given speed using just climb rate and speed without knowing drag/thrust/mass combination?
gripen
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Hi Gripen,
>Well, excess power is not what we are looking for because the propeller is the limiting factor despite what ever amount of power is available.
I guess had should have stated more clearly that the available power curve in my diagram accounts for propeller efficiency, too.
>(except that you appear to use military rating values for the P-51 and WER values for the P-38)
For the sake of the example, I was using the values from F4UDOA's performance sheet.
>So how can you estimate accurate acceleration (value of m/s2) at given speed using just climb rate and speed without knowing drag/thrust/mass combination?
A climb rate for a certain condition is equal to the aircraft's specific excess power for the same condition. Specific excess power has weight (and thereby, indirectly, mass) already figured in:
Ps=P/(m*g)
Newton tells us:
P=Fv; F=ma => a=F/m=(P/v)/m=(P/m)/v
With:
Ps=P/(m*g) => Ps*g=P/m
this results in our acceleration formula:
a=(Ps*g)/v {edited to fix a typo}
Instanteous acceleration at best climb speed at sea level is 2.6 m/s^2 for the P-38J, compared to 1.9 m/s^2 for the P-51B at its best climb speed.
This is absolutely exact, but (unfortunately) not terribly useful since the speed of best climb differs :-)
Here's where the estimate that specific excess power at low speeds is fairly constant comes into play:
The P-51B can be estimated to have 95% of its climb rate at the best climbing speed of the P-38J: At 160 mph, the P-38 accelerates at 2.6 m/s^2 compared to just 2.3 m/s^2 for the P-51.
Likewise, at 200 mph, the P-38 accelerates at 2.0 m/s^2 compared to 1.9 m/s^2 for the P-51.
I'll also add in the F4U-1, which seems to have its best climb speed at 135 knots and accordingly accelerates at 2.1 m/s^2 at 160 mph and 1.7 m/s^2 at 200 mph.
Regards,
Henning (HoHun)
[ 01-05-2002: Message edited by: HoHun ]
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Wonderful discussion, guys.
Couple of observations -
The P-38 reportedly achieved virtually identical climb rates at everything from 140 mph to 180 mph.
I have read that the F6F achieved its best ROC at 140. However, the high AoA at that speed led to engine airflow problems and overheating, which forced testing and performance verification to be done using a climb speed of 160. This resulted in lower ROC numbers than it should have (and was) capable of achieving. Not sure if the F4U had similar cooling problems at its best climb speed, but it seems reasonable that 135 is slow both from a cooling and a torque management standpoint.
I offer both points as cautions that the starting lines aren't always where we think they are.
Dwarf
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Thanks for that, Dwarf. Interesting post.
The P-38 pilot's manual gives best climb speed as around 160 mph for ferry climb (2,300 rpm & 35" MP) and 180 mph for combat climb (3,000 rpm & 54" MP) at sea level for the H/J/L models.
[ 01-05-2002: Message edited by: Guppy ]
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guppy, what is the weight listed for the 38H in the take off and climb chart?
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Originally posted by HoHun:
a=(Ps*g)/v
I don't like to question your equation, but is it really stated in the correct form?
I believe, as written, the equation says that when g = 0, a = 0 ???
Even more, it says that if g is negative you decelerate, and that if g is greater than 1 you accelerate faster.
Clarify please.
Dwarf
[ 01-06-2002: Message edited by: Dwarf ]
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I believe that g ~ 9.81 m/s^2 in Ho-Huns equation. A constant, that is.
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Hi Dwarf,
>I believe, as written, the equation says that when g = 0, a = 0 ???
As Sagefin pointed out, g is a physical constant, the standard gravitational acceleration of 9.81 m/s^2.
g=0 would imply the absence of gravity, which would make the concept of "climb" pretty difficult due to the absence of a concept of "up" ;-)
But if you'd be flying on the Mars with lower gravity, the same aircraft would have a higher specific excess power in the same flight condition (assuming an atmosphere identical to that of the earth). However, level acceleration would be identical to the value achieved under identical conditions here on earth. Specific excess power already accounts for gravity.
I admit that I should have explicitely defined the meaning of g. It's not the current "G-rate", which I'd note as "load factor n".
Load factor does of course have an impact on acceleration as well - "unloaded" dives with n=0 result in better acceleration due to reducing induced drag to a minimum. The interesting thing is that the benefit of "unloading" is greatest at low air speeds (which would require a large angle of attack to sustain straight flight) - at higher air speeds, the benefit of unloading pretty much vanishes.
Regards,
Henning (HoHun)
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bolillo_loco,
The P-38H in the climb chart portion of the manual is listed with a base weight of 16,100 lbs, corresponding to a military power climb rate of 3,500 fpm at sea level. Best climb speeds (IAS) are given as 178 mph combat and 151 mph ferry at sea level, and do not vary with weight (data provided for weights of 16,100, 18,100 or 19,500 lbs).
The P-38J/L are listed with somewhat higher best climb speeds--180 mph combat and 160 mph ferry at sea level.
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"but a closer look to the AHT's data revealed that there is an error in the ammo loadings for the P-47D, ammunition should weight about 1057lbs not 664lbs"
There was two standard loadouts for the P-47 (Although varied a lot in the field): 267rpg or 425rpg.
Daff
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Originally posted by HoHun:
Hi Dwarf,
But if you'd be flying on the Mars with lower gravity, the same aircraft would have a higher specific excess power in the same flight condition (assuming an atmosphere identical to that of the earth). However, level acceleration would be identical to the value achieved under identical conditions here on earth. Specific excess power already accounts for gravity.
Regards,
Henning (HoHun)
Gravity or load factor?
Isn't the real defining variable for Ps drag?
I'm feeling my way here, but it seems to me that acceleration involves more than simply power to weight. And that climb and accel. while similar, aren't really equivalent.
To maximize climb, you put the aircraft in the attitude that maximizes L/D and go to full power. To maximize acceleration, you put the aircraft in the attitude that minimizes total drag and go to full power. Not the same attitude. Not the same drag. And not really the same value for Ps either.
Max accel in level flight would be yet another drag state, and a third different value for Ps, no?
Dwarf
[42 edits later I *think* I've said what I mean as clearly as I can - sheesh]
[ 01-06-2002: Message edited by: Dwarf ]
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Hi Dwarf,
>Isn't the real defining variable for Ps drag?
The definition is
Ps=Pe/W=Pe/(m*g)
with
Pe excess power, W weight, m mass and g gravitational acceleration.
Pe is a complex value determined by many factors including drag, but instantaneous acceleration and climb rate are directly interdependend as outlined above.
>Max accel in level flight would be yet another drag state, and a third different value for Ps, no?
Ps is defined for a specific flight condition, so in level flight at the defined speed, Ps for acceleration is the same as for climbing.
"Unloading" is a different flight condition that would give slightly different results, but comparing different airframes, the change of the flight condition will have a very similar effect, so that acceleration in unloaded flight will have the same relation as acceleration in 1 G flight.
It's certainly safe to say that an aircraft with a 10% climb rate advantage will have an advantage in unloaded dives, too.
Remember that aircraft weight varies with fuel loading anyway - we don't really need to care much about variations in the 5% magnitude since a WW2 fighter may easily be 10% lighter on low fuel than on full fuel anyway.
Regards,
Henning (HoHun)
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HoHun,
Very neat indeed! I tried it for couple spreadsheet systems I made some time ago and your system appear to work well. I have used traditional calculation ie:
Drag at given speed
Fd = ― * s * vē * A * Cd
In the case of the P-51B at sealevel and 200mph (from various RAE and NACA papers):
s = airdensity at sea level = 1,225kg/m3
v = 89,4m/s
A = drag area = 2,265m2
Cd = generic Drag coefficient for complete plane at Cl=0,2 and below mach 0,6 = 0,2
Thrust at given speed
Ft = e * Pp/v + Pe
e = efficiency generic = 0,8
Pp = engine power from propeller same as you used 1510hp = 1109850W
Pe = generic exhaust thrust 120kp = 1177N
Useable thrust for acceleration:
Ft-Fd = 8891N
Acceleration:
8891N/4219kg = 2,1 m/s2 (WER results 2,3 m/s2)
Well, it's pretty near your values, anyway it was propeller efficiency why I gave up with this system at past. At sea level this system is pretty accurate, I can estimate even top speed quite well. But at above 10k this system appear to be more and more unaccurate and overall this system need very good data which is rare.
gripen
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Originally posted by HoHun:
Hi Dwarf,
Ps=Pe/W=Pe/(m*g)
with
Pe excess power, W weight, m mass and g gravitational acceleration.
Regards,
Henning (HoHun)[/QB]
This would seem to account for inertia, but not drag. And, inertia isn't the whole story either.
I just don't see how Ps can be the same for both climb and accel, when those two states have, as they must, different Drag values.
After all, it's the increase in Drag more so than the decrease in Thrust that reduces Ps to 0 at some point. Thrust decreases linearly while Drag increases exponentially with increasing velocity.
Plus, you can't determine Ps until you define the desired outcome. Do I want to climb, or do I want to accel? Whichever it is, I must first create the necessary drag state before I can find out whether I even have any Ps, let alone what its value might be.
I would agree, that at the margins, Ps would be the same for both climb and accel, but for the vast middleground between Stall (actually, sitting still) and Max Level Speed, I don't believe that identity would hold.
Dwarf
[ 01-06-2002: Message edited by: Dwarf ]
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I just don't see how Ps can be the same for both climb and accel, when those two states have, as they must, different drag values.
You are measuring an instantaneous acceleration at a certain speed. The drag and thrust at that speed for 1G flight is the same, whether the plane is climbing or in level flight.
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Gripen -
It's propellor efficiency that has consistently defeated me too. All of the definitions (equations) I can find are circular.
Thrust depends on prop efficiency which depends on velocity which depends on Drag and Thrust. :eek:
I could never seem to quit chasing my tail, and like you I have resorted to using a generic 80% for prop efficiency and call it good enough for government work. But I'm far from satisfied with the results.
Dwarf
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Originally posted by wells:
You are measuring an instantaneous acceleration at a certain speed. The drag and thrust at that speed for 1G flight is the same, whether the plane is climbing or in level flight.
Only if you are willing to accept whatever climb rate falls out of merely firewalling the throttle. Because if you don't adjust pitch, you will always climb and not accel. And, even then, you won't climb at your best rate.
If you seek to maximize either climb rate or rate of accel, you need to first adjust your flight condition so as to permit a successful outcome.
Also, in order to reach any meaningful conclusions about relative performance of aircraft, we need to base the discussion on persistent conditions and not instantaneous ones, IMO.[edit] (except that persistent is an oxymoron... gaaaaaa :mad: English is just so inadequate sometimes. :D My gut says drag and inertia and maybe the way you hold your mouth all play a part, but proving it....)
Dwarf
[ 01-06-2002: Message edited by: Dwarf ]
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To illustrate what I'm trying to get at -
Assume a heavily laden buff.
It staggers into the air. If it gets cleaned up and keeps its nose down it has enough Ps to accelerate. But, no matter how squeaky clean the pilot makes it, it does not yet have enough Ps to climb.
Why? Because to climb it must increase the lift is is generating and that also increases the drag it must overcome. In both cases, it would be operating at the same 1G load and be subject to the same inertia. In both cases, the equations we have seen so far would yield the same positive value for Ps. However, that Ps number only has any practical value insofar as it allows the aircraft to reach a great enough velocity that it can begin to climb. ie. (or is it eg.?) Ps is great enough to overcome added skin friction drag but it is not great enough to overcome added induced drag.
To recap: For all positive values of Ps, accelerating is an option. But, for only some positive values of Ps is climbing an option.
Thus Ps is merely a number, and its significance depends not on its magnitude, but on the problem at hand. While climb and accel ARE similar problems, that similarity does not extend to equivalence in all cases. Inferring performance in one regime from data developed about the other may lead to inaccuracies. Or so it seems to me.
Dwarf
[ 01-06-2002: Message edited by: Dwarf ]
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It staggers into the air. If it gets cleaned up and keeps its nose down it has enough Ps to accelerate. But, no matter how squeaky clean the pilot makes it, it does not yet have enough Ps to climb.
If it can accelerate in straight and level flight, it can climb.
Why? Because to climb it must increase the lift is is generating and that also increases the drag it must overcome.
Yes, but only momentarily while G load > 1.0 in order to raise the nose to an angle who's sine is (T-D)/W. Then, it's back to 1.0 G, climbing at a steady rate.
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Originally posted by wells:
Yes, but only momentarily while G load > 1.0 in order to raise the nose to an angle who's sine is (T-D)/W. Then, it's back to 1.0 G, climbing at a steady rate.
Indeed. The problem, when you investigate induced drag closely, is that getting to that new AoA can involve a manyfold (manifold?) increase in drag. While momentary, that spike is more than enough to kill you.
Point being that climb requires a higher initial value of Ps than accel. Not because the requirements are different for the steady-state portion of the maneuver, but because Ps, alone, carries you over that drag spike and keeps you alive.
Dwarf
[ 01-06-2002: Message edited by: Dwarf ]
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Hi Dwarf,
>I just don't see how Ps can be the same for both climb and accel, when those two states have, as they must, different Drag values.
Easy :-)
Pe=Ptotal-Pdrag
and
Ps=Pe/(m*g)
So the power necessary to overcome drag has already been subtracted earlier, and all power that's left can be used to either accelerate or climb.
For any specific flight condition, Ps for acceleration and climb is indeed equivalent.
Regards,
Henning (HoHun)
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Originally posted by HoHun:
Hi Dwarf,
Easy :-)
Pe=Ptotal-Pdrag
and
Ps=Pe/(m*g)
So the power necessary to overcome drag has already been subtracted earlier, and all power that's left can be used to either accelerate or climb.
For any specific flight condition, Ps for acceleration and climb is indeed equivalent.
Regards,
Henning (HoHun)
OK. We're getting closer to agreement.
:cool:
I still go back to my point in the post above, though.
Once established in either the climb or accel regime, I think the two problems are indeed equivalent enough to not matter. It's getting from where you start to that established condition that differentiates climb from accel.
Climb can charge a very high entry fee, while accel lets you through the gate for free.
Dwarf
Maybe this will help. Refer to your diagram.
At Stall speed, Ps is very near its maximum but still increasing.
Somehow, Drag is not being properly accounted for in order to generate that number. Ps is more potential than actual at that point. More hope than fact. We won't know until we try to do someting whether we really have the Ps we think we do.
The numbers would lead us to believe that a pilot could do nearly anything he wished. Climb, accel, or do Whifferdils. How could he not? He's got all the excess power anyone could hope for. Yet, as nearly 100 years of flight has conclusively established, if he tries to do anything other than accelerate, he will crash and burn.
Meanwhile, up at the high speed end of the graph, one or more parts of that aircraft may encounter its critical mach speed and its attendant sharp drag rise before Ps decreases to zero. Now, we have a situation where the aircraft can climb, but can no longer accelerate despite what the Ps number says.
Would somebody, please, hurry up and solve Navier-Stokes? :D
[ 01-06-2002: Message edited by: Dwarf ]
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It's propellor efficiency that has consistently defeated me too. All of the definitions (equations) I can find are circular.
Thrust depends on prop efficiency which depends on velocity which depends on Drag and Thrust.
Prop efficiency depends on power, prop diameter, velocity and density, that's it. It can be found with a series of 3 equations.
Equation A:
X = V^3 * PI * density * diameter^2 /(2*Power)
Equation B:
Y = SQRT[(X/3)^3 + (X/2)^2]
Then:
e = (X/2+Y)^(1/3)+(X/2-Y)^(1/3)
That assumes no losses and there's always losses, so in your calculations, you can use 0.8 * e to get results within 2-3% for just about all cases.
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Originally posted by wells:
Equation A:
X = V^3 * PI * density * diameter^2 /(2*Power)
Equation B:
Y = SQRT[(X/3)^3 + (X/2)^2]
Then:
e = (X/2+Y)^(1/3)+(X/2-Y)^(1/3)
That assumes no losses and there's always losses, so in your calculations, you can use 0.8 * e to get results within 2-3% for just about all cases.
One of my sources does things a little differntly. First, defining something he refers to as Power Coefficient.
The same source then defines a Thrust Coefficient in a similar manner and uses those two terms in conjunction with Advance Ratio to define prop efficiency.
That whole shebang looks like this:
Advance Ratio: J = V/(n*D);
Power Coefficient: c(p) = P / ((density*n^3)*D^5)
Thrust Coefficient: c(t) = T / ((density*n^2)*D^4)
And Finally: (prop efficiency) = J * [c(t)/c(p)]
with P = 550 * brake horsepower
D = prop diameter in feet
n = prop rotational speed (revolutions per second)
density = air density in a standard atmosphere at current operating altitude as expressed in slugs per cubic foot. A slug being 14.59 kg or 32.174 lb(mass)
T = thrust (lbs)
The kicker, of course is that you need an accurate value for T to get prop efficiency, and you need an accurate value for prop efficiency to get an accurate value for T.
Whichever set of formulas comes closer to being accurate, both would seem to contain more than enough error as to render Ps very suspect at the times when you need it to be most accurate.
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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wells -
As a test, I plugged a few values into your equations.
D = 10 ft
density = .001869 * 32.174 = .06 (8000 ft for the 3 folks whose eyes haven't already glazed over ;) )
P = 550000 (1000 hp)
For V I used both 250 fps and 300 fps.
Both times, after working the 3 equations thru fully, I arrive at a raw e value of 1.
Applying your fudge factor to that result, I wind up back at the generic 80% efficiency I was already using.
Maybe if I tried more values for V I might eventually get something *very* slightly less than 1 for an answer, but I very much doubt it.
It seems we're no closer to a meaningful prop efficiency than we were before.
Anybody else got a set of formulae that isn't circular (unlike the set I already had)?
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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Dwarf,
If you have a good data set on a propeller, you can build a deterministis model but as you can see, it's a case intensive solution.
gripen
[ 01-07-2002: Message edited by: gripen ]
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Heya's,
Go away for a couple of day's and the thread explodes.
Anyway I see things have gone beyond my level of understanding. However...
I still disagree that climb and acceleration are 100% directly proportionate.
For example. Burt Routan's Voyager has an extreme Aspect Ratio to maximize lift with minimal power. If you took the same design and reduced the aspect ratio to as low as you could and maintained the same power without increasing weight while still being a flyable Aircraft which would climb better and which design would accelerate better?
It is just an extreme example to prove the point I was trying to make. That you can't directly link climb and acceleration.
Here is another question.
Based on Thrust-Drag / Mass how would the same A/C acclerate?
[ 01-07-2002: Message edited by: F4UDOA ]
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For example. Burt Routan's Voyager has an extreme Aspect Ratio to maximize lift with minimal power.
This seems to be where you misunderstand. Lift supports the weight of the plane. It doesn't matter what the aspect ratio is, only what the wing area is. What a high aspect ratio does, is minimize induced drag. As you can see from the climb rate equation, when drag is less, excess thrust is more and the plane climbs better. Or, the power required at cruising speed is less.
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Dwarf, for density, you should be using 0.00237 slug/ft^3
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Wells,
Would an A/C with a measurably lower aspect ratio climb as well as one with a high aspect ratio with wing area, power and everything else being equal?
I thought that a high aspect ratio assist in glide, lift and long range IE. Voyager, U-2, B-52 and P-38.
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Originally posted by wells:
Dwarf, for density, you should be using 0.00237 slug/ft^3
That's sea level density. Not sure how that applies if I'm flying at 8000 ft.
In any case it's just a constant, and though it might change the magnitude of the two halves of e, it won't change their difference. The answer at the end would still be 1.
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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Originally posted by F4UDOA:
Wells,
Would an A/C with a measurably lower aspect ratio climb as well as one with a high aspect ratio with wing area, power and everything else being equal?
I thought that a high aspect ratio assist in glide, lift and long range IE. Voyager, U-2, B-52 and P-38.
A. Maybe.
B. Yes.
The reason A is maybe is that, a high aspect wing will produce more lift per unit of drag and therefore requires less power to achieve flight or any given rate of climb. BUT, for any given AoA the faster you go the more total lift a wing produces. The more lift you produce the more induced drag you produce. So, the high aspect wing would perform best at low speeds. A lower aspect wing, while relatively less efficient, would be inherently faster. So while the high aspect wing might permit a 3k ROC at 120, the low aspect wing might allow a 3k ROC at 180. As long as the low aspect plane has enough power to sustain 180, they would climb at the same rate.
The U-2 is a high altitude wonder, but you wouldn't want to dogfight with it. The wings wouldn't take the stress.
As long as you are willing or able to fly slowly, the high aspect wing will permit long duration missions because they require less power for level flight.
Dwarf
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Hi F4UDOA,
>It is just an extreme example to prove the point I was trying to make. That you can't directly link climb and acceleration.
Here's the explanation for your example:
Ps=Pe/(m*g)=(Pt-Pd)/(m*g)
If you increase the necessary power to overcome drag (Pd) while leaving the power available as thrust (Pt) constant, you have an aircraft with different Ps.
Two aircraft with different wing shapes, but equal thrust and drag in a specific flight state are equal in climb and acceleration.
Climb and acceleration are directly and linearly linked.
Regards,
Henning (HoHun)
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Two aircraft with different wing shapes, but equal thrust and drag in a specific flight state are equal in climb and acceleration.
Climb and acceleration are directly and linearly linked.
HoHun,
I am not an engineer but I do understand some of the more basic equations and principles such as lift, drag, thrust etc. but I rely on engineer types like yourself and Wells to clue me in on the rest. But Somehow I think something is getting lost in the translation.
For instance. I have speadsheets, one done by Wells and another done by Zigrat (whom has not been on the boards in a while) calculating Climb, turn, thrust and drag through out the speed range as long as you know the wing area, Aspect ratio, HP, wingspan and 1G stall speed (clean, power on). It gives a very realistic out come based on input. If I put in the F4U-1 data from the spreadsheet I have posted above I will get a climb rate of approx. 3150FPM at Sea level. However if I change one characteristic of the F4U-1, Wingspan the Aspect ratio increases and the climbrate increases several hundred FPM.
Why??
I will post this Spreadsheet on my Webpage and then supply a link.
Standbye.
BTW Zigrat if your out there I will credit you on my Webpage.
[ 01-07-2002: Message edited by: F4UDOA ]
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Hi Dwarf,
>Now, we have a situation where the aircraft can climb, but can no longer accelerate despite what the Ps number says.
You're trying to compare different flight conditions.
An aircraft very close to its top level speed can either climb or accelerate. If it starts to climb, it still can climb "one second later". The flight condition has not changed*.
(* Neglecting altitude since a one-second climb will make little difference.)
If it starts to accelerate, drag will increase, specific excess power will decrease, and it will hit top speed "one second later". The flight condition has changed.
I think what's confusing you is are the different results of applying specific excess power for climbing or for acceleration for a finite time.
Regards,
Henning (HoHun)
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Hi F4UDOA,
>However if I change one characteristic of the F4U-1, Wingspan the Aspect ratio increases and the climbrate increases several hundred FPM.
Does the spreadsheet contain a column for acceleration? That one should increases proportionally.
Regards,
Henning (HoHun)
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HoHun,
Your saying an A/C will accelerate and climb faster with all other factors being equal if the wingspan is increased?
Why?
Here is the link to the spreadsheet. It is in Excel.
I hope Zig didn't copyright it or anything :)
Performance calculator (http://members.home.net/markw4/ubird.xls)
[ 01-07-2002: Message edited by: F4UDOA ]
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Originally posted by HoHun:
Hi Dwarf,
>Now, we have a situation where the aircraft can climb, but can no longer accelerate despite what the Ps number says.
You're trying to compare different flight conditions.
Regards,
Henning (HoHun)
Not really. All I've specified is that the aircraft has encountered a drag rise due to mach. At that point the pilot has only one option - reduce speed. He can do that by either reducing power, or climbing.
Maintaining his current power and attitude just beats his head against a wall. Further accel just went out the window.
At the point 0.001 fps before he encountered mach, your Ps equation might lead him to believe he had another 20 mph in the bank.
It's a poor predictor. I don't know that any better predictor currently exists, but that doesn't change the fact that (especially) at the margins the tools we currently have are not adequate to the task.
Ps is just a number and there are situations at both ends of the flight speed spectrum where it may lead you to try things that just aren't possible. Especially if you believe that climb and accel are the same problem and are linearly linked.
Their differences at the margins differentiate them. If they didn't (to use a gross and unfair example) the aircraft would launch verically into the air as soon as you start the engine. Instead you must accelerate to a speed where flight is possible. At the low end, you can accel when you cannot climb and at the high end you *may* encounter a situation where you can climb but can no longer accelerate.
(Which is exactly what you see in a terminal speed dive, to use another example.)
Climb and acceleration have very different entry requirements. When you include those entry requirements and not just limit yourself to conditions that obtain after those requirements have been met, an aircraft can see a Ps with a zero or negative value with respect to climb while the same power setting, airspeed, altitude etal will yield a positive Ps number with respect to acceleration. If you ignore the entry requirements, the Ps number will be the same for both climb and accel, but, depending on where you are in the flight envelope, trying the wrong maneuver will fail.
The idea that if it can accelerate it can climb and if it can climb it can accelerate is not borne out by 100 years of flying history.
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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Well, at 1g stall speed the plane flys at Clmax of that speed so certainly it can't climb but it can accelerate ie aerodynamic restriction for climb exists. So basicly this point is different than the max climb rate and max acceleration speed when there is no aerodynamic restriction but available energy can be used for what ever wanted. At max speed all thrust is used to counter drag so if it starts to climb it will slow down until balance is reached again.
Generally HoHuns system is pretty neat and actually it is related to old thrust/drag system but calculation is done backwards. And it also is accurate if there is no huge difference between climbing and accelerating flying condition (and there is not) and used data is accurate.
gripen
[ 01-07-2002: Message edited by: gripen ]
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Originally posted by Dwarf:
Anybody else got a set of formulae that isn't circular (unlike the set I already had)?
The equations you had were perfect, they are not circular, they are intended to be used with a set of prop curves for the specific prop in question. So for example, you could calculate the trust from the equation you posted for the thrust coefficient, you would just need the thrust coefficient curve to do it.
Badboy
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Originally posted by Badboy:
The equations you had were perfect, they are not circular, they are intended to be used with a set of prop curves for the specific prop in question. So for example, you could calculate the trust from the equation you posted for the thrust coefficient, you would just need the thrust coefficient curve to do it.
Badboy
Thanks, Badz.
Now... anybody got prop curves for the props on the AH planes? :D
So, basically, I'm just stuck with my generic 80% without those curves.
Dwarf
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Originally posted by F4UDOA:
For instance. I have speadsheets, one done by Wells and another done by Zigrat (whom has not been on the boards in a while) calculating Climb, turn, thrust and drag through out the speed range as long as you know the wing area, Aspect ratio, HP, wingspan and 1G stall speed (clean, power on). It gives a very realistic out come based on input. If I put in the F4U-1 data from the spreadsheet I have posted above I will get a climb rate of approx. 3150FPM at Sea level. However if I change one characteristic of the F4U-1, Wingspan the Aspect ratio increases and the climbrate increases several hundred FPM.
Why??
Because they are using the wing aspect in their calculations to estimate the lift related drag coefficient, and as you increase the wing aspect, the lift related drag coefficient gets smaller and the acceleration and climb rate increase accordingly.
Badboy
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Originally posted by F4UDOA:
HoHun,
Your saying an A/C will accelerate and climb faster with all other factors being equal if the wingspan is increased?
Why?
Yes. Because the other factors don't stay equal, if you increase the wingspan, the induced drag coefficient is reduced.
Here is the link to the spreadsheet. It is in Excel.
I hope Zig didn't copyright it or anything :)
Firstly, bear in mind, that spread sheet is only doing a generic thrust calculation and also an approximate estimate (using the same Oswald factor for every aircraft) of induced drag, so although the results are good for getting a feel for the behavior and trends involved, it won't actually be correct for any particular aircraft.
Having said that, let me tell you how to edit that spreadsheet to provide acceleration data.
Open the spread sheet and go to the cell R12 and type in the following:
=O12*32.2/(E12*129)
That will give you the acceleration of the aircraft in mph every second.
Basically that calculation takes the climb rate, or Ps displayed in cell O12 and multiplies it by g/v to get acceleration and the 129 is a conversion factor so that the units come out in miles per hour, every second.
Once you edit that cell, copy it, then paste it down the entire column.
Badboy
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Originally posted by Dwarf:
Ps is just a number and there are situations at both ends of the flight speed spectrum where it may lead you to try things that just aren't possible. Especially if you believe that climb and accel are the same problem and are linearly linked.
Ps is a very important number, and it is related to climb rate and acceleration. HoHun is absolutely correct.
The idea that if it can accelerate it can climb and if it can climb it can accelerate is not borne out by 100 years of flying history.
Dwarf, most of your statements here (including those I've snipped) are incorrect. Acceleration and Climb rate are directly related, they only differ by a factor that includes (v/g) and so it is possible to make direct comparisons. Trust me on this, HoHun is correct.
Badboy
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Question: Where on the internet can I purchase 10 more points for my IQ?
Anyway
Badboy!!
Great to hear from you again!
I will edit my spreadsheet ASAP.
Anyway for all parties involved this is the root of my question. Take a F4U-1 in comparison with a P-38J. I know that the P-38 could outclimb the Ubird. But what I do not understand is how a P-38 can outstrip an F4U in acceleration when the Power to weight ratio favors the P-38 but not significantly and the Cdo heavily favors the F4U as well as top speed at sea level.
Based on this I can only believe that the Aspect Ratio of the P-38 assist in climb. But the larger wing and increased drag would counter any power to weight advantage.
What factors make an A/C superior in climb/Acceleration?
[ 01-07-2002: Message edited by: F4UDOA ]
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Originally posted by F4UDOA:
Anyway for all parties involved this is the root of my question. Take a F4U-1 in comparison with a P-38J. I know that the P-38 could outclimb the Ubird. But what I do not understand is how a P-38 can outstrip an F4U in acceleration when the Power to weight ratio favors the P-38 but not significantly and the Cdo heavily favors the F4U as well as top speed at sea level.
Based on this I can only believe that the Aspect Ratio of the P-38 assist in climb. But the larger wing and increased drag would counter any power to weight advantage.
What factors make an A/C superior in climb/Acceleration?
[ 01-07-2002: Message edited by: F4UDOA ]
I'll take a stab.
Higher aspect = less power needed at any AoA because the wing generates less drag at that AoA. Thus, more of the potential power the engine(s) can create is available to both climb and accel. Other things being equal, the aircraft with the higher aspect wing will climb faster and accelerate faster.
However, at the other end of the spectrum, the high aspect wing will also run out of acceleration potential sooner.
Ignoring critical mach for the moment, there is only so far you can reduce AoA before you start creeping up the back side of the drag curve. Because the low aspect wing must operate at a greater effective AoA at all speeds in order to produce enough lift to maintain level flight at that speed, it will still have accel potential left when the high aspect wing runs out of gas.
Thus, the 38 climbs and accels better, but the Hog is capable of a higher top speed.
Dwarf
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DOA, the drag coefficient is not a constant. It includes the induced drag, which varies with speed and lift coefficient. Those drag coefficients are for maximum speed, where acceleration is 0 and induced drag might be < 10% of the total. At 150 mph, the drag coefficient of the F4u might be 0.04, with induced drag being 50%. Dwarf, those prop efficiency formulas are very accurate, even with the 20% estimate of losses. I'll show you as soon as I put some graphs up.
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Originally posted by Badboy:
Acceleration and Climb rate are directly related, they only differ by a factor that includes (v/g) and so it is possible to make direct comparisons. Trust me on this, HoHun is correct.
Badboy
I've never said the two problems aren't related or similar. What I have said is they are not equivalent. Equivalence implying point for point identity.
You yourself say they "differ by a factor that includes (v/g)". 1 != 1.0001.
What I've tried to emphasize is that climb and acceleration differ most markedly at the margins. That is where you need to be most wary of the Ps value. Ps is a value which is most accurate and believable for both problems precisely when it is needed least and most unreliable when it is needed most - at the margins.
At low speed you can safely believe it with respect to acceleration, but not climb. At high speed you can safely believe it with respect to climb but not accel. In between, it doesn't matter, because inertia alone will carry you through either transition state. In fact it's inertia alone which allows you to recover from a terminal velocity dive. At terminal velocity, Ps is most likely negative. If that truly meant what it implies recovery would not be possible.
Where I was in error, was early on, before I'd really dug into the problem, believing that climb and accel were most similar at the margins and most different everywhere in between.
Believe me, I wish Ps was more useful. I'd love for it to be. It would make performance modeling and prediction so much easier. I'd love for wells' equations to have been dead accurate at calculating prop efficiency. Or even mostly accurate. I really want something that is better than simply assuming an efficiency of 80%.
I'd also like equations that didn't require graphs I don't have. Purely mathematical means, if accurate, are much more useful to me. BTW, my source does include generic graphs for a "typical 3 blade propellor". I'm unwilling to try to extrapolate much from those for any specific 3 blade propellor or any 2 or 4 blade prop.
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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Dwarf, the number of blades matters not, only the diameter. The number of blades determines the diameter (or vise-versa). The formulas I gave you for prop efficiency should give you this graph for the F4u-1.
(http://www.iaw.com/~general6/f4u_prop.jpg)
Then, you can figure out thrust and drag
(http://www.iaw.com/~general6/f4u_force.jpg)
From there, you can figure out climb rates, glide ratios, sustained turning speeds, whatever you want. Those formulas predict the F4u to have a best climb speed of 70 m/s (136 knots), which is what? 1 knot error from what's in the manual?
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Originally posted by wells:
Dwarf, the number of blades matters not, only the diameter. The number of blades determines the diameter (or vise-versa). The formulas I gave you for prop efficiency should give you this graph for the F4u-1.
...
From there, you can figure out climb rates, glide ratios, sustained turning speeds, whatever you want. Those formulas predict the F4u to have a best climb speed of 70 m/s (136 knots), which is what? 1 knot error from what's in the manual?
Here's what my source has to say.
"For a two-bladed propeller, the forward-flight efficiencies are about 3% better than shown... but static thrust is about 5% less than shown... The reverse trends are true for a four-bladed propeller. Also, a wooden propeller has an efficiency about 10% lower due to its greater thickness."
It's the "about" that troubles me. Fudge factors, especially approximate ones, don't make for very reliable calculations.
The definition I have for prop efficiency says that it is "the ratio of thrust power obtained to energy expended". In fact, another equation for prop efficiency listed in conjunction with that definition is:
n(PE) = Pt/delta(E) = 2 / ( u/u0 +1)
u = propwash velocity
u0 = freestream velocity
The problem with this one, for me, is it requires you to assign values to u and u0. Thus, by adjusting those values you can skew the result to anything you desire whether any actual prop could operate at that efficiency or not.
Make 'em the same and you've got a 100% efficient prop and no thrust. Go the other way and you've got a Hollywood wind machine. But, nowhere, do you have any assurance that your efficiency number corresponds to reality. :eek:
Dwarf
[ Hope Badz didn't read this as originally posted. That WAS pretty erroneous. :o ]
[ 01-07-2002: Message edited by: Dwarf ]
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Take a look at this site:
http://beadec1.ea.bs.dlr.de/Airfoils/propuls4.htm (http://beadec1.ea.bs.dlr.de/Airfoils/propuls4.htm)
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Originally posted by wells:
Take a look at this site:
http://beadec1.ea.bs.dlr.de/Airfoils/propuls4.htm (http://beadec1.ea.bs.dlr.de/Airfoils/propuls4.htm)
Thanks. Excellent site. Just wish some of his equations didn't come with caveats.
Dwarf
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Originally posted by F4UDOA:
1. The P-38J has two 1600HP engines and has a VMAX of only 415MPH.
6. The P-38J has a relatively short range in comparison to other A/C. Why? I thought the P-38 had longer legs. Looking at the P-38 Manual it would seem that it consumed a great deal of fuel compared to the F4U and only carried marginally more fuel.
Any comments?
V Max for the 38 was due to compressibility. IIRC, one early model was really capable of 437, but they were all restricted to lower speeds due to the fear of losing aircraft and pilot at higher speeds.
I believe the listed ranges (radius of action) are predicated on operation at approx 2/3 throttle, auto rich, and the aircraft's best cruising speed. At lower power settings and leaned out mixtures they were all capable of considerably greater range than "the book" figures. On occasion, with drop tanks, there are documented 38 missions of over 2,000 miles.
They used the 38's to get Yamamoto because nothing else in the theater had the legs.
38 compares well with the F7F, which if I'm making out the numbers correctly carried the same 360 gallons.
Even in those days, the government wasn't about to document or encourage modes of operation that had been deemed unsafe. It's at least probable that all US aircraft could exceed their book numbers in some respect. you always need to look at whether the data originated with the manufacturer or relevant service. Manufacturers tend to be optimistic while the services tended to be conservative.
Dwarf
[ 01-07-2002: Message edited by: Dwarf ]
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Gents,
Zigrat has just reared his head from the depths of Aeronautic Engineering school to send me his latest Performance Calculator.
New Calculator (http://www.iit.edu/~buonmic/airtest3.xls)
I think the results of these calculations somewhat prove my point. If you look at the difference in calculated performance no where in the climb numbers does the P-38 exceed the F4U by more than 100FPM.
This is not saying that the F4U should be able to climb with a P-38. Mainly because when a radial engine A/C climbs the clowl flaps are opened created extra drag. However in straight line acceleration this condition does not exist. And the numbers are almost identical.
P-38J
Weight= 16500LBS
Wingspan= 52FT
Wing Area= 327Sq ft
Max Speed @ Sea level= 338MPH
1G stall @ gross weight= 106MPH
F4U-1
Weight= 12000LBS
Wingspan= 41FT
Wing Area= 314 Sq FT
Max Speed @ sea level= 355MPH
1G Stall @ Gross Weight =100MPH
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Dwarf,
Two quick things.
1. How does the P-38 compare well with F7F? The F7F is nearly 60MPH faster at sea level where there is no danger of compressabilty. I'm not sure if I understand what you are comapring?
2. I know the P-38 has a reputation for long range but when the Navy's range calculator is applied it is not nearly as impressive. 450 miles in combat range with no DT's. I know during the course of the war many changes were developed to run lean to get longer range. However this calculator gives a baseline for range for all A/C with out leaning out the mixture. I would just expect more based on reputation.
Also in regard to the Yamamoto mission. I wouldn't put to much stock in "It was the only A/C for the Job". More decisions where based on interservice rivalies and politics than on what was the best for the Job.
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Hi Dwarf,
>V Max for the 38 was due to compressibility.
Though the P-38 had a lower critical Mach number than most contemporary fighters, it still was unable to reach it in level flight.
Regards,
Henning (HoHun)
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Hi Dwarf,
>At low speed you can safely believe it with respect to acceleration, but not climb. At high speed you can safely believe it with respect to climb but not accel.
Your low speed example is misleading because you implicitely assume that initially, the plane is in level flight and has to pull up to enter a climb. You're right it can't pull up while flying at 1 G stall speed without stalling. However, if the initial situation is a climb, this is not necessary, and aircraft can climb quite well at 1 G stall speed.
Your high speed example is slightly flawed, I'm afraid. At top speed, an aircraft uses all of its power to overcome drag, and there's no power left to climb. Only by going slower, the aircraft can begin to climb - just like it could accelerate by descending.
The math isn't that complicated:
Ps=Pe/(m*g)=(Pt-Pd)/(m*g)
=> acceleration: a=(Ps*g)/v, climb rate: Vv=Pe/W=Pe/(m*g)=Ps
The conclusion:
a=Vv*g/v
Acceleration and climb are directly and linearly interdependend.
The derivation of the above formula requires no knowledge of aerodynamics at all, it's a fairly simple application of Newton's axioms.
Regards,
Henning (HoHun)
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my sheet will not work well for the p38 since it is kind of stacked against it. the fact that the nacelles block a large portion of the spanwise flow will mean that the p38 generates less induced drag but the sheet does not take this into account.
as for aspect ratio, yes in general a larger aspect ratio would lead to higher climbrates given equal area, especially at higher altitudes or low speeds. What you must remember though is that high aspect ratio is costly in terms of wing areal weight, so there is a point of diminishing returns where the increased structural weight penalty eliminates the benefits due to reduced induced drag. Its a similar thing to winglets... sure they may decrease spanwise flow, but they add weight and parasitic drag so they cannot be used indiscriminately.
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also wells uhmm those equations you have up there dont work. they always equal unity.
as for number of blades not mattering thats not true. a single blade propeller would be the most efficient actually -- as number of blades is increased efficiency decreases. of course you may able to be absorb more power, but efficiency is still lower.
other things matter too, like design lift coefficient of the blades, taper ratrio of the blades and millions of otehr things. its just that where the hell can you find the activity factor of a bf-109 propeller? damned if i know.
the hamilton standard red book has great data for propeller performance calculation, its just that you cannot find data on the propellers themselves anywhere.
anyways for anyone who wants to learn about aircraft performance the best text book i have found is Jan Roskams Airplane Aerodynamics and Performance (ISBN: 1884885446). It is pricey but is the best book i have found on the subject because anyone can understand it but it still has all the information you need.
all in all tho these little algebraic equations we use have to be taken with a grain of salt. the stuff i am using now in my graduate work for analysis takes weeks to run on sun workstations and still makes assumptions. they are still valuable as estimation tools, but really cant be used for much more than that.
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Hi F4UDOA,
>I think the results of these calculations somewhat prove my point.
I'm afraid your message isn't entirely clear on which point exactly they prove.
If I plug in the values for the P-38J, Zigrat's spreadsheet predicts a climb rate of less than 3300 fpm compared to the 3730 fpm listed by the comparison chart you posted, so my conclusion would be that Zigrat's spreadsheet may be inaccurate.
Regards,
Henning (HoHun)
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Originally posted by wells:
Dwarf, the number of blades matters not, only the diameter.
Hi Wells,
I do see your point here, but the equations are misleading. They do make it appear as though what you are saying is true, but it isn't. The number of blades does make a difference. Prop curves are unique to each type of propeller. If you change the number of blades or their design, you need a new propeller polar. I have just checked some real prop curves for three and four bladed props of an otherwise identical type and the difference in efficiency varies between 6 and 10 percent. Change the prop type to that of another manufacturer and the differences might be even more significant.
You can only use the same prop curve, with any hope of reasonable results if you apply it to only those aircraft that have props with the same number of blades of the same type.
The analogy would be if you were trying to use the same lift curve on every aircraft regardless of the type of wing. Different shape wings, or a different number of wings would require that you use the appropriate curves... Similar thing with propellers.
Hope that helps.
Badboy
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Originally posted by Dwarf:
What I've tried to emphasize is that climb and acceleration differ most markedly at the margins. That is where you need to be most wary of the Ps value. Ps is a value which is most accurate and believable for both problems precisely when it is needed least and most unreliable when it is needed most - at the margins.
This appears to be complete nonsense.
At low speed you can safely believe it with respect to acceleration, but not climb. At high speed you can safely believe it with respect to climb but not accel. In between, it doesn't matter, because inertia alone will carry you through either transition state. In fact it's inertia alone which allows you to recover from a terminal velocity dive. At terminal velocity, Ps is most likely negative. If that truly meant what it implies recovery would not be possible.
So does this.
Believe me, I wish Ps was more useful. I'd love for it to be. It would make performance modeling and prediction so much easier.
And this.
Badboy
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Originally posted by Zigrat:
my sheet will not work well for the p38 since it is kind of stacked against it.
I was also going to mention that you can't really use the same prop thrust curve fit for both the P-38 and F4U, but I think you covered it :)
Badboy
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Originally posted by Zigrat:
its just that where the hell can you find the activity factor of a bf-109 propeller? damned if i know.
You can work it out yourself if you don't mind spending the time with the propeller (at an aircraft museum) and some measuring equipment.
Badboy
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Originally posted by Zigrat:
my sheet will not work well for the p38 since it is kind of stacked against it. the fact that the nacelles block a large portion of the spanwise flow will mean that the p38 generates less induced drag but the sheet does not take this into account.
This reminds me, when I came to analyze the P-38 I ended up doing almost a complete rewrite of my modeling tools to allow for the twin engine configuration. It was frustrating, because it was a lot of extra work for only one aircraft.
Badboy
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Originally posted by Dwarf:
BTW, my source does include generic graphs for a "typical 3 blade propellor". I'm unwilling to try to extrapolate much from those for any specific 3 blade propellor or any 2 or 4 blade prop.
I think that is the first thing I have seen you say, that actually makes good sense!
Badboy
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Originally posted by F4UDOA:
[QB]Dwarf,
Two quick things.
1. How does the P-38 compare well with F7F? The F7F is nearly 60MPH faster at sea level where there is no danger of compressabilty. I'm not sure if I understand what you are comapring?
QB]
I was comparing range. If I made out the numbers correctly, both carry 360 gallons of fuel and the P-38 actually has a slightly better radius of action.
Dwarf
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Originally posted by HoHun:
Hi Dwarf,
>V Max for the 38 was due to compressibility.
Though the P-38 had a lower critical Mach number than most contemporary fighters, it still was unable to reach it in level flight.
Regards,
Henning (HoHun)
Agreed. I didn't mean to be cryptic, but what I meant to convey was that the documented speed was, at least in the case of some 38's, lower than what it could actually achieve. Intentionally so. The AAF didn't want its pilots trying to go any faster than about 415 in the 38.
From my reading of the history of the plane, the AAF imposed a lot of restrictions on how it could be flown. Some justifiable and some not. My bet is that the documented numbers for the airplane are based at least as much on those restrictions as they are on the planes actual capabilities. Just a hunch.
Dwarf
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Awwww, stop it Badz, yer makin me blush.
That's probably the nicest thing you've ever said about me.
Care to offer some proofs? Or do ya just want to snipe?
Dwarf
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Originally posted by F4UDOA:
Also in regard to the Yamamoto mission. I wouldn't put to much stock in "It was the only A/C for the Job". More decisions where based on interservice rivalies and politics than on what was the best for the Job.
True enough in general, but I doubt this in the case of the Yamamoto mission.
As far as interservice rivalry goes, the Marines were the dominant force on Guadalcanal, not the Army. The P-38 flight leader afterwards said that there was no way his squadron would have been picked if F4Fs or F4Us could have made the trip.
Moreover, the Marine fighter ops exec who helped plan the mission specifically stated that the P-38s were the longest-range fighters available.
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Originally posted by HoHun:
Hi Dwarf,
Your low speed example is misleading because you implicitely assume that initially, the plane is in level flight and has to pull up to enter a climb. You're right it can't pull up while flying at 1 G stall speed without stalling. However, if the initial situation is a climb, this is not necessary, and aircraft can climb quite well at 1 G stall speed.
I agree entirely. Once a climb has been established it can be maintained, even at speeds down to Stall.
And once either climb or accel has been established, Ps will do a useful job of providing rate data.
All I've been trying to do is get across the idea that Ps will NOT tell you whether you can get from your current flight condition to that established state in all cases.
Your high speed example is slightly flawed, I'm afraid. At top speed, an aircraft uses all of its power to overcome drag, and there's no power left to climb. Only by going slower, the aircraft can begin to climb - just like it could accelerate by descending.
The math isn't that complicated:
Ps=Pe/(m*g)=(Pt-Pd)/(m*g)
=> acceleration: a=(Ps*g)/v, climb rate: Vv=Pe/W=Pe/(m*g)=Ps
The conclusion:
a=Vv*g/v
Acceleration and climb are directly and linearly interdependend.
The derivation of the above formula requires no knowledge of aerodynamics at all, it's a fairly simple application of Newton's axioms.
Regards,
Henning (HoHun)[/qb]
Please reread my posts. In the first instance, I specified that some part(s) of the structure had encountered critical mach and its attendant drag rise. Whether the equation(s) we have for deducing Ps accurately would take that drag into account isn't apparent. At that point, the aircraft could no longer accelerate, but it could still climb.
In the second instance, I specified terminal velocity, not merely top speed. The aircraft is in a dive and can go no faster no matter what you do, but it still has enough altitude to effect recovery.
For the sake of discussion, let's take a hypothetical aircraft. It has a top level speed of 400 mph. Due to it's configuration, not even WEP and gravity can get it to exceed 600 mph.
It enters a dive at near top speed. Once it passes above that 400 mph top speed, Ps becomes negative. Soon it reaches its 600 mph terminal velocity. Now nothing can make it go faster, but it can still recover (climb). Not because there is some Ps hiding somewhere, but because of the inertia it carries. So long as the stresses don't cause disintegration, inertia will carry it through the transition into climbing flight, after which it will slow until Ps again becomes positive and it can sustain some rate of climb.
At the point at which recovery is begun, it can climb but it can no longer accel. Ps would say it can do neither.
At all points between (slightly above) Stall Speed and Terminal Velocity, the value of Ps doesn't really matter so far as entering a maneuver goes. Both climb and accel can be initiated at will. Both can be sustained at some rate and for some period of time.
Even at those speeds between top level speed and terminal velocity, the pilot can still accel or climb at will. At such speeds Ps can no longer provide any accel rate information and can only tell you that climbing will reduce your speed. It may assist somewhat as to how much accel is being retarded and decel would occur in the climb, but probably not all that much.
Dwarf
[ 01-08-2002: Message edited by: Dwarf ]
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Hi Dwarf,
>All I've been trying to do is get across the idea that Ps will NOT tell you whether you can get from your current flight condition to that established state in all cases.
Yes, you have to think of a specific Ps value as a snapshot of the situation. Transitions between specific situations involve a range of Ps values - if it were a movie, you'd have to examine Ps for every frame :-)
>In the first instance, I specified that some part(s) of the structure had encountered critical mach and its attendant drag rise.
I didn't miss that, but for the Ps concept, the source of the drag is not of interest. If level top speed is reached, and we look at the movie frame of that instant, we'll see that both acceleration and climb are impossible at that speed. The frame before that, he could have done both - and though the acceleration value might have looked good enough from that speed, the sudden drag rise at compressiblity ate it up in the next frame. Ps didn't tell us in advance, the aircraft had to go there to find out :-)
>In the second instance, I specified terminal velocity, not merely top speed. [...] Now nothing can make it go faster, but it can still recover (climb). Not because there is some Ps hiding somewhere, but because of the inertia it carries.
Now I begin to see your point :-) Of course, this is a transition and accordingly more complex, but if we only look at the zoom climb following the climb-out, we certainly see an aircraft climbing at negative Ps for a moment.
Ps indeed describes sustained climbs and instantaneous accelerations, which looks a bit strange to the untrained eye ;-) But a more complete description is that Ps is the rate at which the sum of kinetic and potential energy changes in a specific flight condition. We've seperated this into pure climb and pure acceleration, but in real life, it's usually a combination of both.
Regards,
Henning (Hohun)
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Although this wan't where I was headed when I started, I seem to have arrived at the idea that a new way of looking at climb and accel needs to be devised.
HoHun is absolutely correct when he says that current practice considers climb and accel to be the same problem.
Since the equations to prove it don't currently exist, it's rather hard to conclusively demonstrate that they are not.
When considering the problem of climb, I believe we need something that accounts for the entry cost as well as the cost of maintenance.
Anybody got any ideas how we could do that?
Dwarf
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Originally posted by HoHun:
Hi Dwarf,
Yes, you have to think of a specific Ps value as a snapshot of the situation. Transitions between specific situations involve a range of Ps values - if it were a movie, you'd have to examine Ps for every frame :-)
...
Ps indeed describes sustained climbs and instantaneous accelerations, which looks a bit strange to the untrained eye ;-) But a more complete description is that Ps is the rate at which the sum of kinetic and potential energy changes in a specific flight condition. We've seperated this into pure climb and pure acceleration, but in real life, it's usually a combination of both.
Regards,
Henning (Hohun)
And, I'm probably expecting too much from Ps ;)
What I want is something that can predict as well as confirm. Of course I'd like to be rich and handsome, too :D
Dwarf
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Originally posted by Dwarf:
I was comparing range. If I made out the numbers correctly, both carry 360 gallons of fuel and the P-38 actually has a slightly better radius of action.
Dwarf
The P-38J in the test has 300 gallons of fuel. Later models had a greater fuel capacity of 410 gallons.
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Originally posted by Dwarf:
Although this wan't where I was headed when I started, I seem to have arrived at the idea that a new way of looking at climb and accel needs to be devised.
Nope, everything you need to know about climb and acceleration is expressed in the idea of specific access power, that's all a fighter pilots needs to make a comparison between the ability to accelerate or climb at any practical load factor.
HoHun is absolutely correct when he says that current practice considers climb and accel to be the same problem.
Since the equations to prove it don't currently exist, it's rather hard to conclusively demonstrate that they are not.
The equations to prove it do exist, they are fairly easy to derive, and have been familiar to aero students for a very long time.
Badboy
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Originally posted by F4UDOA:
Dwarf,
Two quick things.
1. How does the P-38 compare well with F7F? The F7F is nearly 60MPH faster at sea level where there is no danger of compressabilty. I'm not sure if I understand what you are comapring?
2. I know the P-38 has a reputation for long range but when the Navy's range calculator is applied it is not nearly as impressive. 450 miles in combat range with no DT's. I know during the course of the war many changes were developed to run lean to get longer range. However this calculator gives a baseline for range for all A/C with out leaning out the mixture. I would just expect more based on reputation.
Also in regard to the Yamamoto mission. I wouldn't put to much stock in "It was the only A/C for the Job". More decisions where based on interservice rivalies and politics than on what was the best for the Job.
A couple of points, if I may?
Grumman's F7F was a "hot ship". Very fast, tremendous climb rate and a punch only matched by the P-61. However, it was also a much more demanding aircraft to fly than the P-38. Corky Meyer has described the Tigercat's power-on stall characteristics as, "adventuresome". Likewise, minimum single engine control speed was notably higher. Another point made by Corky was the loss of rudder control a low speeds, whereas the P-38 had, essentially, blown rudders. For the F7F, this became even more critical should one engine need to be shut down. Not so for the P-38. This is exemplified by the large increase in rudder area on the F7F-3. Outward visibility was better for the Lightning as well. One must remember that the F7F had been in development for a very long time, dating back to the XF5F. One of the generally overlooked aircraft of the immediate pre-war period was the Grumman XP-50. If Grumman's performance numbers are to believed, this beauty was capable of 424 mph, and offered a climb rate in excess of 4,000 fpm. Bob Hall was forced to bail out of the prototype after a Turbocharger exploded, doing such damage as to make the aircraft unsafe to land. After the loss of the only existing example, the USAAC pursued another Grumman twin, the XP-65. This, however, was never built. Eventually, the Navy was to benefit from all the data gathered from the XF5F-1, XP-50 and XP-65, incorporating it into the G-51, or XF7F-1. Indeed, the XP-65 was nothing more than a USAAF spec version of the XF7F-1. Like the XF6F-1, both the XP-65 and XF7F-1 were initially intended to use the Wright R-2600 engine. Shortly after the USAAF cancelled their version, Grumman switched to the R-2800.
One test pilot described the Tigercat's performance as being "like a pair of Bearcats bolted together at the wing root." Not faint praise.
As to the P-38 being the only fighter in the South Pacific that could have pulled off the Yamamoto intercept, what else was there? Nothing. Think of the challenge while remembering that this was an 800+ mile overwater flight, with the outbound leg being flown at wave-top height. Then, assuming that they arrived exactly on time (three or four minutes on either side would have resulted in missing Yamamoto's flight), they had to fight off escorts and any Japanese aircraft taking off, shoot down both G4Ms (Bettys) and then fly all the way back. These were not P-38J of L fighters with the extra 110 gallons of fuel in the leading edge tanks. These were old G models which used unbaffled external tanks. Assuming these P-38s took off with a full fuel load, excluding 60 gallons for warm-up and takeoff, the best range they could expect was 580 miles. Now, figure on using Military power for 30 minutes. This consumes as much as 167 gallons. In other words, they would have to fly 400+ miles each way on just 403 gallons. A check of the P-38 manual (for the P-38H) reveals that these pilots could expect to get 1.02 miles per gallon of fuel burned. This means that the best they could hope for was enough gas to fly 412 miles each way with 30 minutes of combat. That was really cutting it close.
Now, on paper, the F4F-3 appears to have a maximum range allowing them to fly out far enough. However, in practice, the Wildcat had an operational radius of no more than 300 miles, 500 with external tanks. Not good enough. Furthermore, the Wildcat lacked the speed to evade the escorts or to overtake the Bettys quickly, should they find themselves more than a few miles behind.
No, there was no other type in theater that could have performed the Yamamoto mission.
My regards,
Widewing
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as for number of blades not mattering thats not true. a single blade propeller would be the most efficient actually -- as number of blades is increased efficiency decreases. of course you may able to be absorb more power, but efficiency is still lower.
Yes, a single blade is more efficient, but only because you can use a larger diameter. As you add blades, the diameter has to decrease for the same power. That's why efficiency is lower with multiple blades.
also wells uhmm those equations you have up there dont work. they always equal unity.
Umm, no they don't. They were cut and pasted from my spreadsheet that gave the graph I posted.
other things matter too, like design lift coefficient of the blades, taper ratrio of the blades and millions of otehr things.
Well, now you're trying to calculate losses to an optimum propeller, which I admit is difficult. But, given the performance of the plane from flight tests, as in this case, such losses can be determined without taking any of that stuff into consideration.
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Originally posted by Zigrat:
all in all tho these little algebraic equations we use have to be taken with a grain of salt. the stuff i am using now in my graduate work for analysis takes weeks to run on sun workstations and still makes assumptions. they are still valuable as estimation tools, but really cant be used for much more than that.
Amen, brother Zigrat! Is there a Hallejulah for the brother? ;)
I'm afraid we all have a tendency to fall in love with our equations. Especially when they "prove" the point we wish to make.
But as Zigrat points out, they all make simplifying assumptions, and provide only estimates. Sometimes questionable estimates.
Dwarf
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wells -
Zigrat's post and my own prior results intriqued me, so I did 4 more tests of your equations.
v = 50 fps; e = 0.806, after applying your fudge factor, e = 0.6448
v = 100 fps; e = 0.116 and 0.0928
v = 150 fps; e = 0.986 and 0.789
v = 600 fps; e = 1.00005 and 0.8
I suspect that if I investigated further, I could pin down the exact point at which the equation stops oscillating and settles down. As of now, it doesn't appear stable until some point between 100 fps and 150 fps. At that point and for the entire rest of the speed range it appears to yield a result very close to unity.
Dwarf
[edit] Tests were done using your preferred density of .00237 slugs/ft^3, same 1000 hp engine and 10 ft prop.
[ 01-08-2002: Message edited by: Dwarf ]
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Hi Dwarf,
>HoHun is absolutely correct when he says that current practice considers climb and accel to be the same problem.
>Since the equations to prove it don't currently exist, it's rather hard to conclusively demonstrate that they are not.
The equations I posted above prove that climb and acceleration are directly and linearly linked.
The usefulness of the concept is demonstrated by the "energy compensating variometer" used by sailplane pilots - it effectively is a Ps indicator. Sailplane pilots couldn't afford to rely on flawed concepts :-)
>When considering the problem of climb, I believe we need something that accounts for the entry cost as well as the cost of maintenance.
You're asking for energy, not for power now. You could numerically sum up P over short time intervals (the frames in our film) to arrive at the desired solution. (Mathematically, you'd have to integrate.)
Regards,
Henning (HoHun)
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Originally posted by HoHun:
Hi Dwarf,
The equations I posted above prove that climb and acceleration are directly and linearly linked.
Prove is too strong a word in my book. Your equations certainly say they are "directly and linearly linked". However, like all currently existing equations, they contain simplifying assumptions that gloss over parts of the problem that have, so far, been opaque to analysis.
Probably the truest statement we could make is, "From the parts of the picture that have so far been revealed, it appears that the two problems are linked."
The usefulness of the concept is demonstrated by the "energy compensating variometer" used by sailplane pilots - it effectively is a Ps indicator. Sailplane pilots couldn't afford to rely on flawed concepts :-)
...
You're asking for energy, not for power now. You could numerically sum up P over short time intervals (the frames in our film) to arrive at the desired solution. (Mathematically, you'd have to integrate.)
Regards,
Henning (HoHun)
The variometer part I can see. Whether Ps is truly "energy compensating" is the question.
As you point out, climb is a problem that is only amenable to integration. Acceleration is soluble with the Ps formula you have supplied. That seems a significant difference to me.
Dwarf
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wells -
v = 100 fps; e = 0.116 and 0.0928
I suggest you re-calculate that one. There's no oscillating going on...it's a smooth curve .
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Originally posted by wells:
I suggest you re-calculate that one. There's no oscillating going on...it's a smooth curve .
Right you are. Forgot to do the sqrt for Y.
Corrected result: v = 100; e = .959 and .768
Dwarf
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Ok Gents,
While everybody is checking to see who has the biggest protractor. Riddle me this...
Why does the F6F-5 have a higher climb rate than the F4U?
F4U-1
HP=2250
Weight 12,000LBS
Top speed at sea level= 355MPH
1G stall speed @ gross weight = 100MPH
Wing Span 41FT
Wing area 314Sq FT
F6F-5
HP =2250
Weight= 12500LBS
Top speed @ sea level = 330MPH
1G stall speed @ gross weight= 90MPH
Wing Span = 42FT
Wing Area= 334Sq FT
Using these numbers the F6F should not be able to break 3,000FPM but somehow it reaches 20,000FT in 7.0 minutes with an initial sustained climb rate of almost 3500FPM.
BTW. If you look at the test reports of the F6F-5 vrs the A6M5, and FW190A5 and F4U-1D the F4U out climbed the Hellcat in both test. This is not the case in AH.
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Hmmmm... I always thought it was an issue with the F4U-1, not the F6F-6.
It was my understanding it was a propellor efficency problem, that was later corrected on the F4U-4.
Just from memory tho :)
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Heya Verm,
Your right on the Prop efficiency thing. It was with the F4U-1. The blade design was changed midrun to 6501A-0 while maintaining a three blade prop. This change to a paddle type prop is largely documented in the P-47 but nothing is said of the F4U change except in test report and the flight manual. This increased the climb of the F4U significantly.
My question to the engineer types is why the lack of acceleration in the F4U. At first I tried using the P-38 as a comparison but the twin engine makes the calculation come out wrong. So the F6F is a logical comparison because of weight, HP and wing area being so close. Power to weight ratio favors the F4U as well as Drag.
So why does the F6F climb/accelerate better than the F4U?
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Hi Dwarf,
>Prove is too strong a word in my book.
My equations are a formal proof in the true sense of the word.
The proof could be flawed, but the only way to disprove the direct connection of climb and acceleration is to do the math and find the flaw.
People with the formal education should appreciate that verbal arguments are entirely pointless now that we have a formal proof up there. I admit that for anyone without this formal education, I probably don't seem to make much sense :-) The point is: You don't even have a one in a billion chance to be right unless you find a flaw in either my assumptions or in my math.
So this is the point where you'd either have to accept the connection I proved, disprove it formally, or state that you believe I'm wrong whereupon we could both move on to further aspects or the performance comparison :-)
Regards,
Henning (HoHun)
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While everybody is checking to see who has the biggest protractor. Riddle me this...
Why does the F6F-5 have a higher climb rate than the F4U?
Maybe my protractor isn't big enough, but your chart says the F6f climbs at 3160 fpm initially.
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Originally posted by Dwarf:
But as Zigrat points out, they all make simplifying assumptions, and provide only estimates. Sometimes questionable estimates.
Dwarf
In the case of the Ps equation, the difficulty only arises when you attempt to evaluate it. You might choose to introduce simplifying assumptions at that stage, as Zigrat has done in his spreadsheet by assuming lift and drag coefficients that are constant with mach number, as just one example. But that was his choice, the Ps equation itself is exact and conclusions drawn from it are sound.
Badboy
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Originally posted by Dwarf:
Prove is too strong a word in my book. Your equations certainly say they are "directly and linearly linked". However, like all currently existing equations, they contain simplifying assumptions that gloss over parts of the problem that have, so far, been opaque to analysis.
The Ps equation can be found in almost every aerodynamics text book, with all the associated mathematics. Nothing is glossed over or opaque. The equations are simple and crystal clear.
Probably the truest statement we could make is, "From the parts of the picture that have so far been revealed, it appears that the two problems are linked."
The simple truth is that the whole picture has been revealed and they are linked, absolutely and fundamentally.
As you point out, climb is a problem that is only amenable to integration. Acceleration is soluble with the Ps formula you have supplied. That seems a significant difference to me.
Both climb and acceleration can be solved for using the Ps equation. For a fighter pilot there is no real difference, because positive specific excess power means that he has the ability to climb or accelerate or both.
Badboy
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Originally posted by HoHun:
So this is the point where you'd either have to accept the connection I proved, disprove it formally, or state that you believe I'm wrong whereupon we could both move on to further aspects or the performance comparison :-)
Henning (HoHun)
Or, as more usual with Dwarf, he will continue to offer specious argument, ad nauseam :-)
Badboy
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Originally posted by HoHun:
Hi Dwarf,
So this is the point where you'd either have to accept the connection I proved, disprove it formally, or state that you believe I'm wrong whereupon we could both move on to further aspects or the performance comparison :-)
Regards,
Henning (HoHun)
It seems my meaning in an earlier post got misunderstood.
What I meant by: "Since the equations to prove it don't currently exist, it's rather hard to conclusively demonstrate that they are not.", is that the formal proofs are all on your side.
Could we agree that climb is a special case of the broader acceleration problem?
Dwarf
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Heya Wells,
I wasn't speaking of your protracter. However there seems to be a minor conflict going on hear that is somewhat off topic.
Anyway my chart and Zigrats Spreadsheet numbers do give a climb rate of 3160 Per minute. However if you look at the AH charts they show a climb rate of nearly 3500FPM.
My question is based on the numbers used in AH more than IRL. I already have data showing otherwise such as the F4U-1D/F6F-5 vrs the A6M5 and the FW190A5 vrs F4U-1D and F6F-3.
Since you are involved in another simulation what do you feel is the best means of FM modeling.
1. FM based on calculation such as Zigrat's and your spreadsheet.
2. Based on test data such as NAVAIR docs and test trials documents.
3. Based on anecdotal evidence. Such as testamony from pilots.
Just curious.
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dont use my stuff and expect any type of accuracy greater than 90%. The fidelity just isn't there and I don't want people running around using those numbers as bible. Now if something is like 50% off of what it calculates or something, then mabye something is up. If we could predict airplane performance with an excel spreadsheet i wouldnt have to school for four years to concentrate in one discipline of aerospace engineering.
as for dwarf, the other two are correct. Ps translates directly to either climb or acceleration, they are interchangable. thats why Ps vs mach number and altitude is such a valuable plot, since it shows both climb rates at any gives speed, maximum speed versus altitude, and aircraft ceiling. Its basically everything you would want to know about flight at a single level of g.
As for why the f6f climbs better, I don't know. Mabye something about the airplane? The propeller or the cowl flaps? Dunno.
Just to reiterate please don't use that excel spreadsheet as a tool and expect accurate results. The prop efficiency stuff is purely fictional, as are all of the other curves I have seen here. To say that they can be aoppplied to any ww2 airplane is kind of silly.
Plus there are all sorts of factors that this type of analysis dooesnt consider. The fact that the wings are blown by the propellers, stuff like cowl flaps opening and closing based on throttle setting, variations in lift/drag due to camber, thickness, twist, airfoil selection, and planform. The list goes on and on.
Basically I just made up that sheet for fun, and to see if HTCs numbers were in the ballpark. They are.
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Hi Dwarf,
>Could we agree that climb is a special case of the broader acceleration problem?
Climb and acceleration are manifestations of the broader specific excess power complex.
Regards,
Henning (HoHun)
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Hi F4UDOA,
>Why does the F6F-5 have a higher climb rate than the F4U?
>[...]
>Using these numbers the F6F should not be able to break 3,000FPM but somehow it reaches 20,000FT in 7.0 minutes with an initial sustained climb rate of almost 3500FPM.
The Navaer charts list a combat power climb rate of 2980 fpm at 12740 lbs and 2000 HP for the F6F-5. (Time to 20000 ft is listed as 7.7 min.) However, I think this might be without water injection - at 2250 HP power as listed in the above chart, I'd expect the climb rate to be about 3500 fpm. Unfortunately, I can't confirm which power setting the 60" Hg boost listed in the above chart refers to.
Regards,
Henning (HoHun)
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Originally posted by HoHun:
My equations are a formal proof in the true sense of the word.
The proof could be flawed, but the only way to disprove the direct connection of climb and acceleration is to do the math and find the flaw.
Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.
The problem is that during your climb, you gain altitude. This means your best climbspeed changes. Stay at constant TAS and your drag changes. Stay at constant IAS (usual procedure) and you have the problem that TAS becomes higher during your climb. This means your plane accelerates during your climb, and a (small) amount of your power is also used for this acceleration. So your climb performance is in a dynamic climb a bit lower with constant IAS compared to a static climb calculation (simple excess power calculation) what doesn't take this effect into account.
BTW Hohun, your drag power curve in your chart is wrong. You seem to neglect induced drag. Drag, as a force, has a v^-2 characteristics at slow speeds (induced drag) and a v^2 characteristics at high speed (zero lift drag). This means for a power (F*V) the characteristic should be v^-1 near stall speed and v^3 at high speed. But your curve is flat (v^0) at slow speed and ~v^2 at high speed, so it's either a drag FORCE (neglecting induced drag) in a POWER chart, or its wrong.
niklas
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Originally posted by niklas:
Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.
Not climbspeed, climbrate! And that is not wrong, horizontal acceleration can easily be converted into a climbrate. Naturally the situation is dynamic, as you point out, but the relationship still holds even though the speed may be changing, and still works for any particular instantaneous airspeed.
The problem is that during your climb, you gain altitude. This means your best climbspeed changes. Stay at constant TAS and your drag changes. Stay at constant IAS (usual procedure) and you have the problem that TAS becomes higher during your climb. This means your plane accelerates during your climb, and a (small) amount of your power is also used for this acceleration. So your climb performance is in a dynamic climb a bit lower with constant IAS compared to a static climb calculation (simple excess power calculation) what doesn't take this effect into account.
None of that makes the slightest bit of difference to the validity of the conversion, because the Ps equation compensates for changes in speed as they occur. Here is the Ps equation in another form:
Ps = v/g * Vdot + Hdot
Where Vdot is the rate of change of velocity with time, and Hdot is the rate of change of altitude with time.
You should be able to see from that equation that variations in one term, automatically effect the others so that if you evaluate it at any particular moment, you will always get a useful and meaningful answer, and you can always determine what either the climb rate, or acceleration might be from any particular Ps value, or specific excess power. If you set Hdot = 0 you can find the acceleration, and if you set Vdot = 0 you can find the climbrate. Alternatively, you can determine what the climbrate would be for any given current acceleration, or visa versa.
Hope that helps.
Badboy
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Hi Niklas,
>Actually you're right when you say power will determine acceleration and climb, but you're wrong if you think your horizontal acceleration can be easily converted into a climbspeed.
I'm sure you noticed the bullet-proof proof up there :-)
>The problem is that during your climb, you gain altitude. This means your best climbspeed changes.
If altitude changes, so does Ps. Ps is specific to the precise flight condition, and the connection of climb and acceleration is a fact as long as you compare them for the same precise flight condition. If you don't observe this, the results will seem like a contradiction at first, but they really aren't.
>BTW Hohun, your drag power curve in your chart is wrong.
It's correct, but I still think you're a good observer :-) The left border actually is not the stall limit, but the Clmax limit so you would see the behaviour you described if I hadn't simply omitted it from the graph as uninteresting.
Regards,
Henning (HoHun)
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Originally posted by HoHun:
Hi Dwarf,
Climb and acceleration are manifestations of the broader specific excess power complex.
Regards,
Henning (HoHun)
:D :D :D
I LIKE it.
Dwarf
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Originally posted by Badboy:
Here is the Ps equation in another form:
Ps = v/g * Vdot + Hdot
Where Vdot is the rate of change of velocity with time, and Hdot is the rate of change of altitude with time.
You should be able to see from that equation that variations in one term, automatically effect the others so that if you evaluate it at any particular moment, you will always get a useful and meaningful answer, and you can always determine what either the climb rate, or acceleration might be from any particular Ps value, or specific excess power. If you set Hdot = 0 you can find the acceleration, and if you set Vdot = 0 you can find the climbrate. Alternatively, you can determine what the climbrate would be for any given current acceleration, or visa versa.
Hope that helps.
Badboy
Some folks may be more used to seeing the equation in this form:
Ps = dh/dt + (V/g) * (dV/dt)
Much as it will surprise him to read it, I agree with Badz here. This equation (in either form), is somewhat more accurate.
Applying this formula to the P-38 and F4U climb/accel example WAY up-topic gives results of:
Climb -
F4U = 59 ft/s (meaning it has to fly 59 feet in order to climb 52 feet) making its actual airspeed 233 ft/s.
P-38 = 70 ft/s and airspeed 271 ft/s.
Accel -
F4U = 8.43 ft/s^2
P-38 = 8.75 ft/s^2
If you prefer something more familiar, simple Trigonometry gives an equally valid answer to the climb problem.
When you climb, you actually fly the hypoteneuse of the triangle defined by both the horizontal distance you cover and the vertical distance you cover. Climbrate just tells part of the story.
Dwarf
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Dwarf,
Hm... AFAIK the speed discused here have been all the time airspeed (IAS or TAS, not ground speed) so there is no reason for angle correction. There might be a little difference in the planes angle of attack (Cl and therefore also in the drag) during full power climb or acceleration depending on case and therefore also in the IAS/CAS correction, but those difference are small and errors caused by these are also small at the best climb rate speed.
gripen
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Hi Dwarf,
>If you prefer something more familiar, simple Trigonometry gives an equally valid answer to the climb problem.
I'm afraid you're making no sense at all.
>Ps = dh/dt + (V/g) * (dV/dt)
If you're accelerating with no change of alitude, dH/dt = 0. If you're climbing at a constant speed, dv/dt = 0.
Regards,
Henning (HoHun)
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Ps = v/g*Vdot +Hdot
that formula does indeed mean what i tried to say. I only have a problem with the term Ps because it looks like a power. But it is a climbrate, actually itīs the climbrate in a non-dynamic, static observation.
Iīm used to the expression ((F-W)*v)/mg instead of Ps (F thrust, W drag)
W in a level flight for a given speed is not W in a climb at the same speed, so while the formula is correct you still canīt determine from a level fight acceleration your exact climbrate.
Hohun, CLmax should define your stall speed.
niklas
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Originally posted by Dwarf
Much as it will surprise him to read it, I agree with Badz here. This equation (in either form), is somewhat more accurate.
Dwarf
The only thing that surprises me is your, apparently endless, capacity to post complete nonsense.
Badboy
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Hi Niklas,
>Hohun, CLmax should define your stall speed.
Actually, the wing keeps providing lift at angles of attack somewhat higher than the one yielding Clmax ("critical" angle of attack). Though this is nice since it gives the pilot a margin of error before actually stalling, beyond critical angle of attack he'll get less lift at more drag in that region of the flight envelope.
In other words, he doesn't really want to go there :-) However, complete stall of the wing with wingdrop or even spin occur only after critical angle of attack has been exceeded by a certain margin that depends on the specific wing design.
Regards,
Henning (HoHun)
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Originally posted by niklas
that formula does indeed mean what i tried to say. I only have a problem with the term Ps because it looks like a power.
That's because it is a power, P sub s (Ps) is the term used for specific excess power, that's the quantity we are discussing.
W in a level flight for a given speed is not W in a climb at the same speed, so while the formula is correct you still canīt determine from a level fight acceleration your exact climbrate.
Of course you can, if you are in level flight, regardless of acceleration, your climbrate will be exactly zero! I do see your point though, if you enter a climb, things will change, and of course so will the thrust and drag values in that equation. So the fact remains that you can always determine your instantaneous climbrate and acceleration from that equation. Naturally, it will be different for every flight condition, and even when you are established in a climb, the climbrate given by the Ps equation will only be correct for the instant at which the values were taken, that's why it isn't very helpful to consider isolated values. To be useful you really need a map of the Ps contours over the entire envelope. That way you can get a true impression of the behavior of the aircraft. In the same way that with a little experience, it is possible to obtain a 3 dimensional image of a hill by looking at contours on a map, it is also possible to obtain a similar perspective from an aircraft's Ps contours, or curves. Some of the work I'm doing involves a three dimensional approach to aircraft performance comparison, so instead of looking at the Ps curves or contours, I compare three dimensional surfaces, a bit like looking at a hilly terrain. It's exciting stuff :)
Badboy
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Originally posted by F4UDOA
While everybody is checking to see who has the biggest protractor. Riddle me this...
Why does the F6F-5 have a higher climb rate than the F4U?
I believe I can offer you some insight into this, if you would care to take it email. Drop me a note and we can talk about it.
Badboy
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Originally posted by F4UDOA
Since you are involved in another simulation what do you feel is the best means of FM modeling.
1. FM based on calculation such as Zigrat's and your spreadsheet.
2. Based on test data such as NAVAIR docs and test trials documents.
3. Based on anecdotal evidence. Such as testamony from pilots.
Just curious.
The best method would probably involve all three, where the first two are used together combining as many sources as possible and some statistical analysis, with the third used for validation and bias. Also, there are other methods that can be used, but unfortunately, regardless of the methods chosen, any expectation of results better than 10% either way would probably be naive. That's why there will always some controversy involved.
Badboy
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Originally posted by HoHun
Hi Dwarf,
I'm afraid you're making no sense at all.
>Ps = dh/dt + (V/g) * (dV/dt)
If you're accelerating with no change of alitude, dH/dt = 0. If you're climbing at a constant speed, dv/dt = 0.
Regards,
Henning (HoHun)
And whichever you're doing, this formula will work to determine the power required. Just eliminate the surplus term. If you're climbing, the formula becomes Ps = dh/dt + V/g and if you're accelerating the formula becomes Ps = VdV/gdt.
If we set dh/dt = Hdot, and dV/dt = V dot, it's the same equation Badboy posted.
Dwarf
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Originally posted by gripen
Dwarf,
Hm... AFAIK the speed discused here have been all the time airspeed (IAS or TAS, not ground speed) so there is no reason for angle correction. There might be a little difference in the planes angle of attack (Cl and therefore also in the drag) during full power climb or acceleration depending on case and therefore also in the IAS/CAS correction, but those difference are small and errors caused by these are also small at the best climb rate speed.
gripen
OK, assuming you want to make the 227 fps velocity (in the case of the F4U) IAS while flying the hypoteneuse of the triangle, then you can still use Trig to give you your new ground speed during the climb. (221 fps).
Either way, the climb in question requires a Ps of 59 fps to sustain. You don't get a 52 fps ROC by only spending 52 fps worth of power.
Dwarf
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Dwarf,
Nonsense, the starting values for these calculations are known ROCs at known airspeeds ie steady climb rate at steady airspeed. Of course you can calculate ground speed but there is no sense att all to mix grounspeed in. I quess you are just trying to continue until others give up.
gripen
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Originally posted by Dwarf
If you're climbing, the formula becomes Ps = dh/dt + V/g and if you're accelerating the formula becomes Ps = VdV/gdt.
Nope, the first expression you posted assumes a climb with acceleration equal to unity. A climb with no acceleration would just give dh/dt. Also dv/dt is a single term, so the second expression you posted is just meaningless nonsense.
If we set dh/dt = Hdot, and dV/dt = V dot, it's the same equation Badboy posted.
You don't have to set dh/dt = Hdot because they are both well known notation (courtesy of Leibniz and Newton respectively) for the same thing.
Badboy