Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: Stoney on July 05, 2010, 02:19:14 PM
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Gents,
Ok, for the last few months, I've been working on a master spreadsheet that catalogs a large number of aerodynamic characteristics for most of the Aces High plane-set, minus stuff like the C-47, for example. Anyway, so far, I've included both the zero-lift and induced drag coefficients, but have yet to include any correction for compressibility drag (that's planned for later). I've put all of this information together so that I can do some quick analysis whenever I want. I started testing out my propeller efficiency formulas and have hit a snag, as I don't really trust the numbers.
For example:
F6F-5 at 25% fuel = 11358 lbs
Wing Area = 334
Clean stall = 96 mph IAS (tested in game)
dynamic pressure = 23.57
Cdi =.1363
Cdoi =.0272
Drag = Cd * 1/2pv^2(dynamic pressure) * Area
Induced Drag @ stall = (.1363)*(23.57)*(334) = 1073.2 lbs
Zero-Lift Drag @ stall = (.0272)*(23.57)*(334) = 214.1 lbs
Total Drag = 1287.3 lbs
Thrust = Drag
Thrust at stall = 1287.3 lbs (not including exhaust thrust)
Power at stall = 550*2000bhp = 1100000
Velocity = 96mph*1.467 = 140.8 ft/sec
Prop Effi = Thrust*Velocity / Power
Prop Effi(stall) = .1648 ~ 16.5%
Now, I know propeller efficiency starts getting wonky when velocity drops off, so I wasn't really sweating this one. But when I plug everything in for max speed, my efficiency number looks pretty whacky. Can the exhaust thrust make a big enough difference to throw the number off this much?
~Max speed at Sea Level = 320
Dynamic pressure = 261.9
Cdi = .0011
Cdo = .0272
Induced Drag = 96.6 lbs
Zero Lift Drag = 2379.3 lbs
Total Drag = 2745.9 lbs
Thrust = 2745.9 lbs
Power = 1100000
Velocity = 469.4 ft/sec
Prop Effi = 1.17 ~ 117% <-------I know that is way wrong. Where did this get screwed up?
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Prop Effi = 1.17 ~ 117% <-------I know that is way wrong. Where did this get screwed up?
Yep, your Cdo value is too large.
Badboy
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Hmm, that number came out of Dean, but I suppose he could have been mistaken. I found another number of .0211. I plug that one in and get:
Prop Effi @ stall: 15.8%
Prop Effi @ 320: 83%
Still seems like the full speed number may be a bit high, but certainly closer to what it should be. Anyone know a good correction for exhaust thrust?
Thanks for the correction...
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Exhaust thrust can be as high at 200lb on some planes.
HiTech
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Found this on the NASA reports server: http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090014842_2009014224.pdf
Looks like they got ~4.5% of total thrust at 325 mph from the experimental data, with the collector ring and nozzle. I'm not sure if the in-service nozzles were as efficient as the test nozzle. Would this be a conservative correction for thrust or is it too different to be used as a valid approximation?
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Anyone know a good correction for exhaust thrust?
Hoerner gives a rule of thumb formula for exhaust thrust:
T [lb]= (0.11 to 0.13)*P [hp]
This does not apply with turbocharged engines, where (I think) you can assume the exhaust thrust as zero.
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Hoerner gives a rule of thumb formula for exhaust thrust:
T [lb]= (0.11 to 0.13)*P [hp]
This does not apply with turbocharged engines, where (I think) you can assume the exhaust thrust as zero.
Is that theoretical max or max realized?
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Ok, another problem I'm having is that the efficiency, if extrapolated beyond Vmax, just keeps increasing to beyond 100%. This doesn't follow most prop efficiency curves I've ever seen. It should be decreasing as it approaches Vmax, right?
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I have assembled this which turned out as I expected it would. Vmindrag is = to Vy, correct?
(http://i125.photobucket.com/albums/p61/stonewall74/F6F-5Drag.png)
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I'd be suspicious of any prop efficiency that comes out lower than about 0.5 or so for an airplane in flight.
Here's a way to estimate prop efficiency using advance ratio and power coefficient, which are calculated from aircraft velocity, prop diameter, prop RPM, and engine HP. See the "Thrust Equations" section of this document:
http://electraforge.com/brooke/flightsims/aces_high/stallSpeedMath/turningMath.html
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Is that theoretical max or max realized?
Neither. I said "Rule of thumb".
IMO that means "You can use this for calculations if you don't have any better data".
I have seen exhaust thrust charts for BMW801 and Jumo 213.
Brookes' formulas seem to be "rule of thumb" -sort also. They don't take account of the blade numbers, blade width, blade airfoil etc.
The equation 2.2 seems to need some iterative calculation.
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Neither. I said "Rule of thumb".
IMO that means "You can use this for calculations if you don't have any better data".
I have seen exhaust thrust charts for BMW801 and Jumo 213.
Brookes' formulas seem to be "rule of thumb" -sort also. They don't take account of the blade numbers, blade width, blade airfoil etc.
The equation 2.2 seems to need some iterative calculation.
Yep, there are more-complicated eta calculations, but the equation will get in the ballpark.
Equation 2.2 isn't iterative, but requires solving a nonlinear (cubic) equation, or using numerical methods or just graphing it (Figure 3-17).
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I have assembled this which turned out as I expected it would. Vmindrag is = to Vy, correct?
Did you mean: at Vmindrag you are climbing at Vymax? If so I would say yes.
For a C47 would be fun to estimate the max angle of climb (very close to stall's angle) if someone wants to sneak a base placed on a steep hill eheh..
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Here's a way to estimate prop efficiency using advance ratio and power coefficient
Using advance ratio is typically only used for determining static thrust. The method I used is normally used for approximating thrust in the normal flight regime. Most designer interpolate the transition area between takeoff and stall by drawing a curve between the two. At least, that's what Dan Raymer says in Aircraft Design: A Conceptual Approach
Now, that being said, I may have made a mistake in calculation, so if you see one, feel free to point it out. Regardless, thanks for the help.
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Neither. I said "Rule of thumb".
IMO that means "You can use this for calculations if you don't have any better data".
I have seen exhaust thrust charts for BMW801 and Jumo 213.
Brookes' formulas seem to be "rule of thumb" -sort also. They don't take account of the blade numbers, blade width, blade airfoil etc.
The equation 2.2 seems to need some iterative calculation.
Well, you're certainly getting that from a "bible", but I was simply curious as to how the exhaust configuration would change that number. For example, a collector ring and pipes (such as on a radial) versus the paired exhausts of the Merlins.
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Okay Stoney your the leader. I'll march 100 miles for you just don't ask me to do the math. And I thought Geometry was hard.
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Prop Effi @ stall: 15.8%
You should be looking for another mistake, because the prop efficiency at stall will be much higher and probably around the 60% mark. At first glance I would double check your induced drag values.
Badboy
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Brookes' formulas seem to be "rule of thumb" -sort also. They don't take account of the blade numbers, blade width, blade airfoil etc.
The equation 2.2 seems to need some iterative calculation.
The equation for prop efficiency can be derived from f=ma and leads to a cubic which has a relatively simple solution that gives the theoretical maximum thrust possible, similar to the curve in figure 3-17 on Brooke's page.
If any one is interested in seeing how the equation is derived and the solution, with an example, I'd be happy to post it.
Badboy
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You should be looking for another mistake, because the prop efficiency at stall will be much higher and probably around the 60% mark. At first glance I would double check your induced drag values.
Badboy
Ok, to back up a bit...
I computed e, K, and Cdi using the following formulas...
e = 1.78(1-.045A^.68)-.64 with A in this case representing Aspect ratio. So e for the F6F-5 = .88 (this formulas was presented by Raymer as a more accurate approximation than the Glauert or Wessinger approximation)
K = 1 / pi*A*e or K = pi*5.5*.88 = .066
Cdi = K*Cl^2
Clmax at stall is 1.44 @ sea level, 11358 lbs and 96 mph IAS. 23.57 for dynamic pressure there.
So...
Cdi = .066*1.44^2 = .1363 at stall.
Di = (1/2pv^2)*S*Cdi or 23.57*334*.1363 = 1073.16 lbs. of drag.
One of these formulas must be wrong, I guess, as I've double checked the math manually. Perhaps Raymer's "e" approximation is bad. .88 seems a bit high, but the math works. I know that .7 to .85 is considered the normal range. According to this link: http://karoliinasalminen.wordpress.com/category/oswalds-efficiency-factor the e-range presented by various authorities varies between .6 - .9. Even with a .6 entered for e, the prop efficiency only rises to 22%.
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The equation for prop efficiency can be derived from f=ma and leads to a cubic which has a relatively simple solution that gives the theoretical maximum thrust possible, similar to the curve in figure 3-17 on Brooke's page.
If any one is interested in seeing how the equation is derived and the solution, with an example, I'd be happy to post it.
Badboy
I'd enjoy seeing that.
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One of these formulas must be wrong, I guess, as I've double checked the math manually. Perhaps Raymer's "e" approximation is bad. .88 seems a bit high, but the math works. I know that .7 to .85 is considered the normal range. According to this link: http://karoliinasalminen.wordpress.com/category/oswalds-efficiency-factor the e-range presented by various authorities varies between .6 - .9. Even with a .6 entered for e, the prop efficiency only rises to 22%.
Raymer's equation is good. It's basis DATCOM emperical data. I'll have a look at your numbers as well.
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Stoney- checking your calcs I think your drag is right but for the stall speed power required you're using the wrong amount of power. 2000hp is the power available. Basis your total drag at stall of 1287 lbs that's 329hp power required. Using an prop eff of .6 the power required is 549 bhp. Hope that helps!
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Stoney- checking your calcs I think your drag is right but for the stall speed power required you're using the wrong amount of power. 2000hp is the power available. Basis your total drag at stall of 1287 lbs that's 329hp power required. Using an prop eff of .6 the power required is 549 bhp. Hope that helps!
Yep, well spotted.
Stoney,
Raymer's equations are ok, as are your calculations, accept for the point dtango made. When you equate thrust to drag you imply an equilibrium (or a steady state) condition, and sustaining the 1g stall speed of 96mph requires much less than full power. As dtango has calculated, if the prop efficiency was 60% at stall, you could sustain that stall speed with 549 horse power. Full power at 1g and 96mph will cause the aircraft to accelerate, so you can't equate thrust to drag under those conditions.
What your calculations show, is that if you really did apply 2000bhp and were only able to sustain 96mph in a 1g stall, the prop efficiency would have to be as low as 16.5%, which is well below what it should be at that speed.
Hope that helps.
Badboy
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Speaking of the prop's in the game why not give us the ability to adjust prop pitch and also put fire extinguisher's in the bomber's engine's so that when the engine catches fire we can put it out :) its a thought it may also improve some of the aircraft. Its a thought like
What you think of it HiTech ????
Be a bit more challenging wont it :)
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also put fire extinguisher's in the bomber's engine's so that when the engine catches fire we can put it out
They won't help.
Engines do not burn in AH.
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Speaking of the prop's in the game why not give us the ability to adjust prop pitch and also put fire extinguisher's in the bomber's engine's so that when the engine catches fire we can put it out :) its a thought it may also improve some of the aircraft. Its a thought like
What you think of it HiTech ????
Be a bit more challenging wont it :)
You can already adjust prop pitch, it's called an RPM control.
HiTech
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Stoneys mistake brought a question to my head.
What is the definition of a power on stall speed?
I see 2 possibilities.
1. Full power in climb stall speed.
2. Min power required to maintain level flight.
HiTech
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Stoney,
A bit busy right now but do you assume that D=T at stall speed?
Edit: Ah... the others have allready noted this, sorry :aok
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Yep, well spotted.
Stoney,
Raymer's equations are ok, as are your calculations, accept for the point dtango made. When you equate thrust to drag you imply an equilibrium (or a steady state) condition, and sustaining the 1g stall speed of 96mph requires much less than full power. As dtango has calculated, if the prop efficiency was 60% at stall, you could sustain that stall speed with 549 horse power. Full power at 1g and 96mph will cause the aircraft to accelerate, so you can't equate thrust to drag under those conditions.
What your calculations show, is that if you really did apply 2000bhp and were only able to sustain 96mph in a 1g stall, the prop efficiency would have to be as low as 16.5%, which is well below what it should be at that speed.
Ok, the lightbulb came on with that last sentence. So, for example, to say eta p at 320mph is 83% is probably close, since my calculation used 2000bhp for that, and at max level speed, power required would be 2000bhp.
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Stoneys mistake brought a question to my head.
What is the definition of a power on stall speed?
I see 2 possibilities.
1. Full power in climb stall speed.
2. Min power required to maintain level flight.
HiTech
For operational purposes, I'd assume stall speed at full power and in climb, since you're mostly concerned with the stall speed in takeoff configuration/attitude with power on. That's why power-on stalls are part of the PPL syllabus right? Just like they teach power off stalls in the dirty configuration to simulate stalls in landing operations? I don't know how that would dovetail into the pure aerodynamic aspect, but that'd be my argument.
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Ok, the lightbulb came on with that last sentence. So, for example, to say eta p at 320mph is 83% is probably close, since my calculation used 2000bhp for that, and at max level speed, power required would be 2000bhp.
Yes. At max level speed your airspeed is constant which means T=D at full engine power. At stall speed with full power T>D. Stall speed drag=1287 lbs as you've calculated. However at 96mph but full power (2000hp) and assuming eta at .6 your thrust would be 4680 lbs, thus you can't assume T=D at stall speed at full engine power...
...unless of course your eta is a crappy 16% ;).
Tango
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Stoneys mistake brought a question to my head.
What is the definition of a power on stall speed?
I see 2 possibilities.
1. Full power in climb stall speed.
2. Min power required to maintain level flight.
HiTech
All the power-on stall procedures I've seen are some variation of #1 and not #2. But what do I know since I'm not a real flight test engineer or test pilot! :joystick:
Tango
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You can already adjust prop pitch, it's called an RPM control.
HiTech
Ahem iu thought RPM meant you move the throttle forward you get more speed like in your car :confused: :confused: :headscratch: :headscratch:
But if you have incorperated it as all in one control then i getcha :) :D
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Ahem iu thought RPM meant you move the throttle forward you get more speed like in your car :confused: :confused: :headscratch: :headscratch:
But if you have incorperated it as all in one control then i getcha :) :D
You have 2 controls on the WWII planes and AH.
1. Is throttle like in the car, it lets more air into the engine.
2. Is the rpm control that sets the governor RPM. (+ and - keys on the number pad)
Apply throttle = more air = more power = governor changes pitch to keep same RPM.
HiTech
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You have 2 controls on the WWII planes and AH.
1. Is throttle like in the car, it lets more air into the engine.
2. Is the rpm control that sets the governor RPM. (+ and - keys on the number pad)
Apply throttle = more air = more power = governor changes pitch to keep same RPM.
HiTech
Ah cool it just a wee bity different to iL2's control's it get's confusing playing both game's at time's. And i am actualy easily confused at the moment altho Ace's is helping me with my medical condition at the time being
And i thank you and your team for that hitech.
oh and please check the wish list i have wished for a new plane :) :salute :salute
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I'd enjoy seeing that.
Ok, here goes...
We start with Newton's law:
(http://www.leonbadboysmith.com/images/1.gif)
Essentially, the propeller accelerates a mass of air, and in that equation m will be the mass of air and a will be the acceleration of the air and then F will be the resulting force, that is the thrust.
So, what we do is to consider what happens in one second. Firstly, we need to calculate how much air will pass through the prop, and then multiply that by the density of the air so that we have the mass. Then we multiply that by the acceleration of the air. We do that by assuming that the air will form a circular shaft, with the same diameter as the prop. We will use letters from the Greek alphabet and typically a change in any quantity is normally preceded by the Greek letter delta, so delta v will be the change in velocity. We use the Greek letter Rho for the density of the air.
So now we can substitute into #1
(http://www.leonbadboysmith.com/images/2.gif)
That is just the mass airflow through the prop per second, multiplied by the change in velocity during that second. The problem is that we have the term delta v in there and later we will want to eliminate it from the equation, so we need some more relationships.
Now we need to introduce the term for efficiency and we use the Greek letter eta, and efficiency is just output power divided by the input power and the output power is the thrust times the velocity and the input power is the power of the engine, and we express that like this:
(http://www.leonbadboysmith.com/images/3.gif)
rearranging that we get:
(http://www.leonbadboysmith.com/images/4.gif)
Also, the power absorbed by the propeller can be expressed as the thrust times the average velocity through it, like this:
(http://www.leonbadboysmith.com/images/5.gif)
If we equate #4 and #5 and then solve for delta v we get:
(http://www.leonbadboysmith.com/images/6.gif)
(http://www.leonbadboysmith.com/images/7.gif)
Now if we solve #5 for the thrust we get:
(http://www.leonbadboysmith.com/images/8.gif)
Now we can equate that to #2 to get:
(http://www.leonbadboysmith.com/images/9.gif)
Now we can substitute the value of delta v from #7 to get:
(http://www.leonbadboysmith.com/images/10.gif)
That looks complicated, but we can simplify it to this:
(http://www.leonbadboysmith.com/images/11.gif)
and we can rearrange that like this...
(http://www.leonbadboysmith.com/images/12.gif)
That is one of the standard forms of a cubic equation, and you notice that two of the terms in there appear to be the same. So we can simplify this even more by letting:
(http://www.leonbadboysmith.com/images/13.gif)
Then #16 is just:
(http://www.leonbadboysmith.com/images/14.gif)
That's an amazingly simple result for the efficiency of the prop, and it can be solved. All we need to do is evaluate a in #17 and substitute it into the standard solution shown below:
(http://www.leonbadboysmith.com/images/15.gif)
Remember, that's going to give us the maximum theoretically possible efficiency, but there will be losses, so one practical solution recommended by Perkins and Hage is to multiply the result by 0.8587. We will do that here for simplicity, but it is possible to estimate the losses and include them in the calculations.
So, let's do an example using the data from earlier in the thread and calculate the efficiency of the F6F prop at 96mph.
Then we have:
D = 13.08
Rho = 0.0024
P = 2000 x 550
v = 140.8
So now we calculate the value of a:
(http://www.leonbadboysmith.com/images/16.gif)
Now we substitute that into the solution given above, evaluate it and apply the reduction factor recommended by Perkins and Hage as follows:
(http://www.leonbadboysmith.com/images/17.gif)
So we have a prop with 64% efficiency, slightly more than my earlier guess.
Hope that helps...
Badboy
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Excellent Badboy, I appreciate the effort that went into posting that, thanks.
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Ok, here goes...
................
Hope that helps...
Badboy
:aok :aok :aok :aok :aok 5/5!