Author Topic: Climb and acceleration(engineers please)  (Read 1189 times)

Offline F4UDOA

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Climb and acceleration(engineers please)
« Reply #15 on: July 10, 2002, 10:28:23 PM »
Wells,

Everytime I say why does one accellerate better than another I hear "because it climbs better" or why does it climb better? Because it accellerates better. This is not a reason.

Try this.

Forget that we know the climb of any WW2 fighters.

All we know is the weight, wing area, aspect ratio and top speed.

Right off of the HTC webpages

Airplane
#1 F4U-1D
Weight =12,175LBS
Wing Span=41
Wing Area= 314
Aspect ratio= 5.35
HP sea level Mil power= 2,000

#2 P-51D
Weight 9611lbs
Wing span= 37
Wing Area= 233
Aspect ratio= 5.89
HP sea level mil power= 1490

#3 F6F-5
Weight= 12,483lbs
Wing span = 43
Wing area= 334
Aspect Ratio= 5.50
HP at sea level= 2,000

Please show me(anyone) which a/c climb or accelerates the best.
Why?

I have run these numbers through Zigrats spreadsheet about a hundred times and it comes up different than what is generally accepted.

Why?
Why?
Why?
:confused:

Offline wells

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« Reply #16 on: July 11, 2002, 12:12:34 AM »
Start with the basics.   What's the absolute 'maximum' climb ability that we can get from a given horsepower?  1 HP is the ability to raise 550 lbs 1 foot in 1 second.  So, for your 3 examples, we get:

Corsair:

RoC = 2000 * 550 / 12175 = 90 fps

P-51D:

1490 * 550 / 9611 = 85 fps

F6F-5:

2000 * 550 / 12483 = 88 fps

We know that the planes will never reach those numbers because of drag and reduced thrust with speed.

Now, we look at static thrust.  The absolute maximum that we can get from 13' 2" and 11' 2" propellers with the given power is:

F4u and F6f:  9182 lbs
P-51D:  6761 lbs

Of course, you'll never see those kinds of figures due to inefficiencies like drag and other rotational losses, but those apply to all prop planes...

Ok, so now we can look at T/W:

F4u:  0.75
F6f:  0.74
P-51D:  0.70
Now, wing loading and aspect ratio, which will determine lift coefficient for a given speed and thus induced drag.  If we square the wing loading, like squaring the lift coefficient, we can get an idea of induced drag coefficients.  Multiplying the coefficients by the area gives an idea of the induced drag force for a given airspeed.

F4u:  38.8^2 * 314 / 5.35 = 88356.7
F6f:  37.4^2 * 334 / 5.50 = 84942.9
P-51D:  41.2^2 * 233 / 5.89 = 67148.3

These are not actual drag numbers, but more like index values for comparison purposes.  We can say, for example, that the F4u will see 1.3 times more induced drag at a given speed than the P-51D and 4% more than the F6f.  Since the F4u has more than 30% greater thrust over the P-51, we can say that the F4u should improve even more over the 51, but the F6f will gain some or maybe even pass the F4u.  So, yeah, everything points to the F4u being superior to the 51, but we still don't know parasite drag, skin friction or prop efficiency...anything could happen!  Keep in mind that more thrust creates more drag as the propwash goes back over the airframe.  That works in the P-51's favour.  The P-51 is likely to see a greater benefit from exhaust thrust, due to the higher manifold pressures and RPM that it uses.  For a given Cd0, the F4u shows 35% more wing area and drag.  If we assume same Cd0, we can say the 51 and f4u will be about the same speed because the f4u has 35% more thrust too.

Wind tunnel tests might help you narrow down the Cd0, while flight tests will then help to get prop efficiency.

Offline Naudet

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« Reply #17 on: July 11, 2002, 01:43:02 AM »
What i am just wondering about:

Couldnt the differences F4UDOA mentions come from different engine power curves?

I think Wells calculations will be all for SL, right?

And if the test was flown at 10K the relative thrust numbers might differ even more than at SL, cause of different charger gears, prop layouts for different performance characteristics (i.e. P51 for high alt, both F4 and F6 for low to medium alt) and so on.

Offline niklas

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« Reply #18 on: July 11, 2002, 05:35:35 AM »
The military power curve of a F4U-1 indeed shows only 2000hp for sealevel, at 3000ft the power dropped already to 1850hp. ("F4U-1 Brake Horsepower VS Altitude" chart oct.44).
The Merlin or Packard engine gains power up to 8000ft or so with miltary power, peaking out at 1590 hp (?)

Offline hitech

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« Reply #19 on: July 11, 2002, 08:34:25 AM »
F4UDOA, In your question about why f4u climbs poorly , the primary difference in the f4u is prop effency at best climb rate speeds (lots of HP to small of prop for it).

Offline Steven

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« Reply #20 on: July 12, 2002, 10:26:49 AM »
F4U has a "small" prop?   I thought Vought slapped an exceptionally large prop to the aircraft requiring either really long landing gear or a bent wing.  It has same enging as Hellcat and P47 but the F4U has a smaller prop?

Offline F4UDOA

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« Reply #21 on: July 12, 2002, 12:08:26 PM »
Wells,

That is as good an analysis as I have seen. Thankyou. It looks like aspect ratio plays a huge part in determining induced drag.

Hitech,

I agree they could have done a better job with the prop selection from the beggining. However the F6F used the same exact prop blade as the F4U with worse power loading and climbed marginally better depending on which report you read.

Which brings me to my next question.

Aspect ratio is a huge factor in induced drag. The Spitfire used an elliptical wing to offset this. However Niklas posted a document saying that all WW2 fighters has a somewhat elliptical wing and that it didn't make that much difference.

My question is how much difference is there in the wing shape of the P-51, F4U and F6F fro reducing induced drag and what effect does wing taper have??

Offline Badboy

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« Reply #22 on: July 12, 2002, 02:55:42 PM »
Hi DOA,

An interesting thread, and so many excellent responses! I would have responded sooner, but since you have been receiving almost perfect responses from just about everyone, I haven’t been able to add anything useful until now.


Quote
Originally posted by F4UDOA
Aspect ratio is a huge factor in induced drag. The Spitfire used an elliptical wing to offset this. However Niklas posted a document saying that all WW2 fighters has a somewhat elliptical wing and that it didn't make that much difference.

My question is how much difference is there in the wing shape of the P-51, F4U and F6F fro reducing induced drag and what effect does wing taper have??



Firstly, Aspect Ratio is the primary factor in determining the drag due to lift, but the taper or spanwise distribution of area of the wing also has an effect, because if the typically elliptical lift distribution acts upon a wing that has its chord lengths distributed in an elliptical manner, the resulting lift will be distributed almost equally over each square foot along the span. That results in an even and almost constant downwash along the span and thus the smallest amount of trailing vortex, and thus induced drag. For example, a rectangular wing, with a taper ratio of one would result in the same elliptical lift distribution acting on a constant chord length, so that the chords near the tip would be carrying less than their share of the load, while those near the root would be carrying more, which for any given Aspect Ratio results in inefficiency due to the variation in downwash distribution and increased vorticity. For example, a rectangular wing with an AR of 6 would have 5% more induced drag than an elliptical wing of the same AR. At the other extreme, with a taper ratio of zero, that would be a 13% difference. Midway between those two extremes a wing with a taper ratio of 0.5 would have a coefficient of induced drag of only about 1% more than an elliptical wing with the same AR. But it isn’t as simple as that because you can achieve chord lengths that vary elliptically without the traditional symmetrical ellipse shape, and also the elliptical variation in lift can be achieved by twisting the wing, or spanwise variations in the wing section camber. Also the elliptical chord variations we have been discussing are only theoretically true for wings with negligible weight, and has been known to be not strictly true for practical wings since the early thirties. That doesn't answer your question about the specific aircraft mentioned, but it does get you into the right ballpark.

Also I don’t think it has been mentioned yet that the results for the P-51 in the tests you have been referring to are not ideal for comparison because the mustang developed engine problems at 4000ft during the test and had to complete the climb on reduced power. They acknowledge in the report that “the data would not be representative.” That is certainly part of the reason for the disparity along with the problem of averaging the climb rate and acceleration already correctly explained by others.  

Lastly, regarding the original question about acceleration and climb, if you are interested in seeing the mathematical relationship between them derived, just ask.

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Offline F4UDOA

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Climb and acceleration(engineers please)
« Reply #23 on: July 12, 2002, 03:29:45 PM »
BadBoy,



I hoped this thread sounded somewhat intelligent. I didn't want to sound like it was whining.

The P-51 did throttle back during the climb but at least it was the same A/C climbing and accelerating. So the relationship should be the same which is what I was questioning. As far as the actual performance goes the Radial engines were all MAP limited by the use of less than ideal fuel while the Merlin was not. Besides the weights of these A/C were all way off, so any results of actual speed or climb are biased by there weights. I think flying characteristics were the focus of their report.

So I understand your point about span distribution what is better?

Higher span loading or lower??

Here are span loadings for several A/C from a Vought report I have.

In no order.

1. F4U-1D======= 294
2. F6F-5======== 291
3. P-47D========345
4. P-51B======== 251
5. P-38J======== 316


Also what is better with taper ratio?? Higher or lower?
From NACA report

1. P-51    Apect Ratio = 5.89 Taper ratio == 2.17
2. F4U-1  Aspect Ratio=5.30 Taper Ratio == 1.47
3. F6F      Aspect Ratio=5.50 Taper ratio ==  2.00

What does this tell you??

Offline Badboy

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« Reply #24 on: July 12, 2002, 08:06:54 PM »
Quote
Originally posted by F4UDOA

I hoped this thread sounded somewhat intelligent. I didn't want to sound like it was whining.


Certainly not whining, but it looks as though you are heading somewhere with your questions, I’m not sure where, and please forgive me if I’m mistaken, but it doesn’t look to me as though you are simply trying to learn something about aerodynamics, I get the feeling you have a point to make?

Quote

The P-51 did throttle back during the climb but at least it was the same A/C climbing and accelerating. So the relationship should be the same which is what I was questioning.


Yep, but there was no mention of engine problems during any of the other tests, so I assume it was running smoothly for all but the climb test, and that might easily account for the difference.

Regarding the data you posted, if I calculate the lift dependant drag due to the trailing vortex wake based on the aspect ratio, the taper ratio, and the sweep angle, using your data supplemented with some of my own, the calculations reveal an induced drag coefficient for the F4U that remains almost constant up to about M0.35 and of the P-51, F6F and F4U, the F4U has the largest coefficient, with just a little less than 2% more induced drag than the P-51 and only marginally more than the F6F. Is that what you were expecting?

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Offline dtango

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« Reply #25 on: July 13, 2002, 01:08:29 AM »
Badboy:

I think his understanding of the relationship between rate of climb and acceleration is confusing him.  The question is he believes that the data that he's seen shows that rate of climb and acceleration don't correlate to each other as he has been told.

F4UDOA:

You're asking a complex question which requires a complex answer.  You're also misunderstanding the relationship between rate of climb and acceleration.  Maybe the following concepts and relationships will help you in understanding the physics and explain the data that you're seeing.  

EQUATION FOR RATE OF CLIMB:
Let's start by examining climb performance.  For a given aircraft steady state climb is expressed by the following equation:
RoC = (Thrust - Drag) * Velocity / Weight

RATE OF CLIMB IS A FUNCTION OF EXCESS POWER OF AN AIRCRAFT:
To simplify this a little further, since we know Power = Force * Velocity, we can then modify the equation with:
RoC = (PowerAvailable - PowerRequired) / Weight
Where:
PowerAvailable = Thrust * Velocity - (we'll call this Pa)
PowerRequired = Drag * Velocity - (we'll call this Pr)
So in essence rate of climb is deteremined by the EXCESS POWER of an aircraft (difference between Pa and Pr) divided by weight.  There is a lot of complexity behind this relationship!  Let's explore this further.  


PA, PR AND EXCESS POWER VARIES WITH  VELOCITY:
Consider the following figure describing the aerodynamic relationship between Pa, Pr and velocity:


FIGURE 1: POWER-VELOCITY CURVE
 
Figure 1 describes the relationship (T-D)*V relationship in the RoC equation.
 
SO HERE ARE THINGS TO NOTE FROM THIS GRAPH:[LIST=1]
  • Pa and Pr vary with velocity.  Pr specifically is a function of total drag [incidently the u-shaped curve occurs because induced drag is high at low speeds and decreases with velocity while form drag is low at low speeds but increases with velocity].
  • For a given velocity (or a given point on the x-axis above) you have a specific excess power (Pa-Pr) value for that velocity.  In other words the amount of excess power changes with velocity.
  • The best rate of climb occurs at the SPECIFIC velocity where excess power is at it's greatest [ where (Pa-Pr) is at maximum].  In the figure this is noted as Vr/c-max.  SO IN OTHER WORDS BEST RATE OF CLIMB IS A FIXED POINT INSTANCE AT A SPECIFIC CONSTANT VELOCITY.
EXCESS POWER (Pa-Pr) = ACCELERATION IN LEVEL FLIGHT:
If an aircraft is not in a climb and in level flight, excess power translates into acceleration for an aircraft.  This is how acceleration and an aircraft's rate of climb relate to each other.  Excess power governs the rate of climb as well as defines an aircraft's acceleration in level flight.  Maximum acceleration for an aircraft will occur at the speed for the best rate of climb of that aircraft.  NOTE THAT THIS IS ONLY A POINT INSTANT IN TIME, HENCE THE TERM INSTANTANEOUS ACCELERATION.  

2NDLY if you look at figure 1 remember that excess power varies with velocity, therefore as an aircraft accelerates through it's it's envelope of airspeeds THE RATE OF ACCELERATION CHANGES BASED ON CHANGING (Pa-Pr).  Looking at figure 1 note how Pa-Pr is small at the left of the graph, then increases as velocity increases until Pa-Pr reaches some maximum value (velocity for best rate of climb) and then decreases as velocity continues to increase until you reach max level speed.  This entire acceleration profile is then the AVERAGE ACCELERATION of the aircraft, or the acceleration of the aircraft as it goes from one velocity to another.

Here's one of the places that I think you're getting tripped up at.  Best rate of climb = where the rate of acceleration is at the maximum.  This is only a point instant in time and can only be described at that SPECIFIC VELOCITY.  What you are looking at from the report data is a snapshot of AVERAGE ACCELERATION over range of velocities.[/i]

There's more...

PA-PR VS. VELOCITY CURVES DIFFER FROM AIRCRAFT TO AIRCRAFT:
So now we are at the point where we understand and analyze comparisons of rate of climb and acceleration between different aircraft.  The Pa-Pr vs. Velocity curves for each aircraft are a function of their aerodynamic characteristics (as simplified by the RoC = (T-D)*V / W equation).  Let's use for the sake of comparison the F6F-5, F4U-1D, and P-51D as already mentioned above.  With what Wells has already provided as well as knowing some general things about the aircraft in question we can make some rough calculation to give us some guesses at their power vs. velocity curves look like.  Figure 2 is a rough stab at this at SL.


FIGURE 2: DIFFERING POWER-VELOCITY CURVES

So the reason you are seeing the differences in AVERAGE ACCELERATION snapshots of the aircraft in question is demonstrated by these curves.  You can see that at lower speed ranges the F6F probably out accelerates the P-51D (F6F Pa-Pr is greater than the P-51D Pa-Pr).  But then at higher speeds ranges, the P-51D out accelerates the F6F because at higher speeds the Pa-Pr for the P-51D is greater than that of the F6F.

Here's another place that I think you're getting tripped up at.  Each aircraft has it's unique power-velocity curves based on it's aerodynamics which determines varying rates of climb and rates of acceleration over varying velocity.  Rates of climb are not constant but vary with velocity.  This means that if Plane A has a better rate of climb vs. Plane B at a given range of velocities (hence better acceleration at those velocities), you cannot assume that Plane A will maintain that rate of climb advantage over (and out accelerate) Plane B at a different range of velocities.[/i]

Hope this helps.

 
Tango, XO
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« Last Edit: July 13, 2002, 01:10:40 AM by dtango »
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Offline Naudet

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« Reply #26 on: July 13, 2002, 05:11:39 AM »
@dtango: Little of topic, but can you post a Pr/Pa diagram for the FW190D9? Best with three different powers (1750/1900/2100PS).

Offline dtango

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« Reply #27 on: July 13, 2002, 01:09:19 PM »
Naudet:

I'll take a look at it.  I'm thinking you want something that is more of an accurate depiction of Pr/Pa curves for the 190D-9.  Figure 2 above really is a rough stab and meant for illustrating a concept.   To put together accurate Pr/Pa curves, even an accurate estimate is pretty involved.  At any rate I'll look into it and see how far I get and post request for info or findings in a new thread.

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Offline dtango

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« Reply #28 on: July 13, 2002, 03:08:09 PM »
One more thing regarding Aspect Ratio and induced drag, it's a mistake to assume that AR is the primary factor in determining induced drag.  Here's why:

It is true that the induced drag coefficient equation is:

CDi =       CL^2
        -------------------
           pi * e * AR

We also know the following as well...

AR =  b^2
       ----------
            S

and

CL =     2W
       -----------------
        S * rho * V^2

and that

Di = CDi * S * rho * V^2


If you substitute the equations for CDi and CL and AR you'll end up with the following equation for Di in level flight..

Di =           2W^2
       ------------------------
       pi * rho *e * V^2 * b^2

So what does this tell us?  Something interesting:

(1) From an airframe perspective weight and wingspan (b and not AR) are the major contributors to induced drag.  If you want to reduce induced drag, decrease weight or increase wingspan.

quoting Don Stackhouse at DJ Aerotech:
Quote
So what controls induced drag? Essentially it's a function of the amount of lift being made (weight and bank angle), how much air is being influenced by the wing (i.e.: "mass flow"), and how efficiently the wing uses that mass flow. Mass flow depends on the density of the air, the speed of the aircraft, and the wingspan. Note that I did NOT say "aspect ratio". Induced drag is NOT directly a function of aspect ratio. For a fixed wing area, an increase in aspect ratio will improve induced drag, but only because the only way to increase aspect ratio for a fixed wing area is by increasing the span. It's SPAN that controls mass flow, NOT aspect ratio and NOT wing area!


(2) Span efficiency (e) is what Badboy was referring to regarding taper etc.  Span efficiency is partly a function of AR.  Span efficiency is between .7-.9 for most planes.  WW2 Aerodynamic changes such as elliptical wings, rounded wing tips etc. etc. still puts span efficiency in the .7-.9 range.

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« Last Edit: July 13, 2002, 05:10:03 PM by dtango »
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Offline Badboy

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« Reply #29 on: July 13, 2002, 04:35:27 PM »
Quote
Originally posted by dtango


So what does this tell us?



That induced drag doesn't change with altitude... Just kidding, I don't mean to be picky, but you don't have a complete dynamic pressure term because you dropped rho from the denominator of your final expression. Otherwise, well explained.

Badboy
« Last Edit: July 13, 2002, 04:46:43 PM by Badboy »
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