XP-39 II - The Groundhog Day Thread

FLYBOYJ said:

Again, that would be shown on a performance chart that includes temperature and density altitude into the equation. I believe most if not all US WW2 fighters computed at "Standard Altitude" (59F at sea level 29.92"HG).

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Okay. The average high temperature for the hottest month on Guadalcanal is 88 degrees at sea level and average low temperature is 73 degrees.
Average high for the hottest month in Dayton, OH (Wright Field) is 84 degrees, average low is 65 degrees.

How much will speed and climb be degraded at Guadalcanal if the test was done in Dayton?

P-39 Expert said:

Okay. The average high temperature for the hottest month on Guadalcanal is 88 degrees at sea level and average low temperature is 73 degrees.
Average high for the hottest month in Dayton, OH (Wright Field) is 84 degrees, average low is 65 degrees.

How much will speed and climb be degraded at Guadalcanal if the test was done in Dayton?

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Depends on if the test results are corrected for STP which they usually are.

- Ivan.

P-39 Expert said:

Okay. The average high temperature for the hottest month on Guadalcanal is 88 degrees at sea level and average low temperature is 73 degrees.
Average high for the hottest month in Dayton, OH (Wright Field) is 84 degrees, average low is 65 degrees.

How much will speed and climb be degraded at Guadalcanal if the test was done in Dayton?

Click to expand...

You will have to factor in air pressure (as mentioned) and understand DENSITY ALTITUDE

Let's start here;

Density altitude - Wikipedia

In the P-39N flight manual I found this:



Notice the note at the bottom of the chart (my red circle). Unfortunately on the climb data the % for each 10C (50F) above 0C (32F) free air temp (circled in yellow) is missing.

FLYBOYJ said:

You will have to factor in air pressure (as mentioned) and understand DENSITY ALTITUDE

Let's start here;

Density altitude - Wikipedia

What you will find that if an aircraft is sitting at sea level and it's 88 degrees, it's going td fly like it's at 2000'. If you're at 10000' and the OAT is 61 (16C) it will perform like its at 12,500.

This will also affect take off performance.

View attachment 601044

Density Altitude:

• Pressure altitude corrected for non-standard temperature
• Used for performance calculations
  • Formula:
    • Pressure Altitude + (120 x [Outside Air Temperature (OAT) - (ISA Temp)])
  • Example:
    • Pressure Altitude = 600' (as calculated above)
    • OAT: 10°C
  • Calculate:
    • ISA Temp (using standard Lapse rate of -2 degrees C per 1000 ft) is 14° C
    • 600' + [120 * (10-14)]
    • 600' + (-480) = 120'

In the P-39N flight manual I found this:

View attachment 601043

Notice the note at the bottom of the chart (my red circle). Unfortunately on the climb data the % for each 10C (50F) above 0C (32F) free air temp (circled in yellow) is missing.

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From the P-39 L/K 10% for every 20F over 32F

P-39 Expert said:

How much will speed and climb be degraded at Guadalcanal if the test was done in Dayton?

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So to answer your question - you can lose 28% climb time combat load, no head wind according to the way I read the 2nd chart.

Why this is removed from the P-39Q flight manual - I don't know???

ThomasP said:

However, the values for P-39C S/N 40-2988 cannot be correct - at least not for a sustained climb at the power setting and weight listed. My math says that the AC can do 3270 ft/min at 6680 lbs with 1160 BHP. (Don't know if the 3270 as opposed to 3720 is an indication of anything, could just be a coincidence.)

The only way I can get the sustained 3720 ft/min ROC is if the engine is putting out ~1300 BHP, or If you accelerate to Vmax at SL and go into a zoom climb at max power, then divide the height gained before you have to reduce power by the time it took to reach said height. I suspect the latter is what they did.

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When I did the calc with an f of 3.77929, e of 1.56299 & prop eff of 0.81 (Perkins & Hage Fig 3-20b), I got an ROC of 3706 ft/min.
e and f calculated per Prof Rogers "Finding e & f from GPS flight test data." Prof Rogers

Hi ThomasP,

I used Aerodynamics for Naval Aviators. There is a free pdf download at: https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80t-80.pdf .
If you look on page 154, there is a formula for Rate of Climb: RC = 33,000 * [(Pa - Pr) / W], where RC = Rate of Climb (fpm), Pa = power available (hp), Pr = power required for level flight (hp), W = weight (lbs).

Since I had tests and some data, I could plug in a known rate of climb at a known altitude and a known power (hp) and back into the required hp for level flight.

Let's look at the P-39C. At 10,000 feet, the rate of climb is 3,720 fpm. An Allison V-1710-35 makes 1,150 hp at 12,000 feet and the airplane is at a weight of 6,689 lbs. It just so happens we have data for 10,000 feet. Rate of climb is 3,720 fpm. Calculating backwards, the power required for level flight at that altitude is 396 hp. Personally, I'd expect the power required for level flight to go up as the altitude goes up becasue there is less air density, so the wing has to "work harder." By work harder, I mean a slightly higher angle of attack, and that creates more induced drag, requiring more power.

Now, I go to 20,000 feet and I have 1,530 fpm, same weight (for ballpark estimates, anyway). The table says 675 hp from the Allison. The power required for level flight is 365 hp, which is logical to me. Suppose I play P-39 Expert's game and assume the weight dropped by 350 lbs and I'm still at 20,000 feet. Using the same numbers as just above, the expected rate of climb would be 1,614 fpm, a gain of 84 fpm.

Now, let's look at the P-39D. Same engine. At 10,000 feet, the rate of climb shows 2,720 fpm and the rest of the data is the same except 7,525 lbs. I calculate Pr to be 530 hp. First, that is very strange because the two airplanes are so similar. The P-39D should not require 1/3 more power in level flight. Nevertheless, let's go to 20,000 feet. Same 675 hp available, same 7,525 lbs. I calculate Pr to be 378 hp! Somehow, it takes much less power to fly at 20,000 feet in thin air than it does at 10,000 feet in much thicker air?

No way. Something is very strange.

Now, go back to the 10,000 feet for the P-39D and assume about an equal difference between power required for the P-39D as for the P-39C. For the P-39C, the power required changed by about 30 hp between 10,000 feet and 20,000 fet. So, I take the P-39D power required at 20,000 feet ( 378 hp) and add 30 hp (408 hp) and calculate the expected rate of climb at 1,150 hp to be 3,254 fpm. That is significantly different from 2,720 fpm. If I next plug in the achieved rate of climb (2,720 fpm), I see the power avilable was only 1,028 hp and not 1,150.

Coincidentally, the normal power at 10,000 feet for the V-1710-35 at 2,600 rpm is 1,000 hp.

On the whole, it looks to me as if the P-39D test was likely carried out at Normal power and not military power. Again, if I lose the 350 lbs that P-39 Expert is wanting to lose, the expected rate of climb at 1,150 hp would go from 3,254 fpm to 3,413 fpm, a gain of 159 fpm. The differences between the P-39C and P-39D can be propeller, a sour engine, or any of many factors. But, it should NOT climb at only 2,720 fpm at 10,000 feet with 1,150 hp avaialble.

I am using the formula from Aerodynamics for Naval Aviators because most of the formulas were developed and checked for military-type airplanes with military type horsepower piston engines. There's the right way, the wrong way, and the Navy way. But the Navy doesn't make too many aircraft design calculation mistakes.

GregP said:

Personally, I'd expect the power required for level flight to go up as the altitude goes up becasue there is less air density, so the wing has to "work harder." By work harder, I mean a slightly higher angle of attack, and that creates more induced drag.

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Yes, there's an increased induced (and parasite) drag effect due to increased AoA, but isn't that more than compensated for by reduced total drag from the significantly lower air density? I've read AfNA too.

P-39 Expert said:

Okay. The average high temperature for the hottest month on Guadalcanal is 88 degrees at sea level and average low temperature is 73 degrees.
Average high for the hottest month in Dayton, OH (Wright Field) is 84 degrees, average low is 65 degrees.

How much will speed and climb be degraded at Guadalcanal if the test was done in Dayton?

Click to expand...

Properly done, it doesn't matter whether the test is done in Dayton, Guadalcanal, or Shangri La. The test report doesn't quote raw data, it quotes observed performance mathematically corrected to standard atmosphere. Without standard conditions you have no basis for comparison of aircraft at different times and places.
So your P39D pilot taking off from Henderson Field at 88°F, 29.82"Hg, and 95%RH has no reason to expect his tired, patched, high time mount to match the performance numbers printed on his Wright Field test report. Dream on.

Hi XBe02Drvr,

For top speed, you are correct.

But for power required to maintain level flight at Vx, I'd expect the required power to go up as you go higher. The rate of climb test is not a top speed test and is run mostly at whatever best rate of climb airspeed is. I'd say it's somehwere around 1.4 * stall speed and slightly increases as you climb, and the power required to maintain level flight at Vx increases, too, as you climb.

Well now.....lets see if someone who has never flown agrees with you guys or not......................................

FLYBOYJ said:

So to answer your question - you can lose 28% climb time combat load, no head wind according to the way I read the 2nd chart.

Why this is removed from the P-39Q flight manual - I don't know???

Click to expand...

I believe the chart says add 10% to elapsed climbing time for every 20 degrees above 32 degrees. Guadalcanal averages 81 degrees. 81 less 32 = 49 degrees. 49 degrees divided by 20 degrees is 2.4min. Adding 2.4min to 14.5min to 25000' is 16.9min. Normal climb 14.5min averages 1724fpm. Adjusted climb 16.9min averages 1479fpm. Difference is 244fpm.

Still not quite 1000fpm. It's the weight.

P-39 Expert said:

I believe the chart says add 10% to elapsed climbing time for every 20 degrees above 32 degrees. Guadalcanal averages 81 degrees. 81 less 32 = 49 degrees. 49 degrees divided by 20 degrees is 2.4min. Adding 2.4min to 14.5min to 25000' is 16.9min. Normal climb 14.5min averages 1724fpm. Adjusted climb 16.9min averages 1479fpm. Difference is 244fpm.

Still not quite 1000fpm. It's the weight.

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You mentioned 88F in your original post.

You're reading/ using the chart wrong as well as doing the math wrong!!

"49 degrees divided by 20 degrees is 2.4min."

Wrong!

"10% for every 20 degrees"! 49 degrees = 24% NOT 2.4 minutes!

YOU ADD ON 24% TO THE CLIMB TIME!

ALSO

It's broken down in segments and the rate of climb changes. Chart edited in blue and yellow boxes.

P-39 Expert said:

I believe the chart says add 10% to elapsed climbing time for every 20 degrees above 32 degrees. Guadalcanal averages 81 degrees. 81 less 32 = 49 degrees. 49 degrees divided by 20 degrees is 2.4min. Adding 2.4min to 14.5min to 25000' is 16.9min. Normal climb 14.5min averages 1724fpm. Adjusted climb 16.9min averages 1479fpm. Difference is 244fpm.

Still not quite 1000fpm. It's the weight.

Click to expand...

And on this chart it gives you the time! It's not "the weight," it's "DENSITY ALTITUDE." This is why these charts were made!

P-39 Expert said:

I believe the chart says add 10% to elapsed climbing time for every 20 degrees above 32 degrees. Guadalcanal averages 81 degrees. 81 less 32 = 49 degrees. 49 degrees divided by 20 degrees is 2.4min ???. Adding 2.4min to 14.5min to 25000' is 16.9min. Normal climb 14.5min averages 1724fpm. Adjusted climb 16.9min averages 1479fpm. Difference is 244fpm.

Still not quite 1000fpm. It's the weight.

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READ WHAT THE CHART SAYS!



SO - the chart says time to 25,000' at 8400 pounds is 17.1. I'll use your 81F. 81F is 49F warmer than 32F. 10% FOR EACH 20F above 32F so 24% Increase (40F = 20% and then add another 9F = 4%, 24%) to 17.1 = 21.2 Minutes to 25K! This at 8400 Pounds!

17.1 x .24 = 4.104 (24% of 17.1) + 17.1 = 21.20 minutes to 25K at 8400 Pounds!

14.5 + 24% = 17.98 minutes to 25K at 7800 Pounds!

13.0 + 24% = 16.12 minutes to 25K at 7400 Pounds!

Ivan1GFP said:

As has been explained to YOU, Those test results are under standard conditions or corrected to standard conditions.
Today happens to be a reasonably cool Fall day and Temperature is in the low 60's or not too far off standard temperature of 59 degrees Fahrenheit. Most of this Summer was way above that. The average daily temperature on Guadalcanal is 81 degrees.
How many nice cool 59 degree days do you think you will find in the SWPA? What do you think happens when you run the same aeroplane in 110 degree temperatures? Do you think performance gets better?

If you are calling 27,000 feet the new combat ceiling, you are simply making things up.

I haven't made anything up, just quoting an AAF report. Why would a General quote a figure for climb at 100fpm?

You have ABSOLUTELY NO EVIDENCE that the two planes were "virtually identical".
On the other hand, you have quite a lot of evidence that SOMETHING, most likely engine power or thrust was significantly different between the two aircraft. As GregP pointed out, the Power Required for the P-39D didn't even match well with what it was doing in its climb at higher altitudes.

Now how does your 1.2 fpm per pound rule translate to higher altitudes and improved service ceiling????

It translates to 10000'. It will decrease with altitude. Just like all climb rates.

Again, you make the assertion that the planes were virtually identical except you have no evidence that was true. They were as identical (except weight) as the manufacturer could make them. Same contract, same engine, same HP, same airframe. Different internal equipment.
There certainly were external aerodynamic differences such as the .30 cal MG in the Cowl, the .30 Cal MG in the Wings. Yep, those .30s would account for 1000fpm of climb.
I am not convinced that there was any extra length between the C and D models. Agree with you and it wouldn't make any difference anyway.
Most of the other factors you brought up are from a different discussion and not relevant here. None of this is relevant except for the almost half ton of extra weight.

As for the Propeller, Your statement really tells us that you know nothing about propellers. And you are an expert?
Take a look at the propellers on the Curtiss P-40. I believe the majority of them were made by Curtiss Electric and I believe all were 11 feet 0 inches in diameter. The blade designs are quite different and their performance is different. Very slightly different performance. The propeller with the biggest difference was a British Rotol propeller. What are the odds of that propeller being used on an American plane? The propellers used on the Allison engine (like in the P-39) had very small differences.

You have been told that the engine outputs listed in the P-39D test do not match the climb performance and given evidence that you choose to ignore. Both were official performance tests at Wright Field.
Any of these other factors alone would make the planes NOT identical. The two planes were as identical as the manufacturer could make them, except for weight. The difference in climb wasn't from different propellers, atmospheric conditions, symmetrical airfoils, tail fillets, .30cal gun ports etc. It was because one was 836lba heavier than the other.

- Ivan.

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Expand above.

P-39 Expert,

Weight is not quite what you seem to believe it is as a parameter.

For example, a Schweizer 2-33 sailplane and a Boeing 747 happen to share about the same glide angle.

But the Schweizer's best glide speed is 52 mph and the 747 is at about 250 knots, which results in quite an actual descent rate in feet per minute.

This is not straight-line algebra.

You guys feel free to use this. I use it when I'm trying to deal with muppet truck drivers

I think all would be clear if we had a colour picture of a p-39 in flight.

Your wish is my command.



It is in flight, fleeing the water, a task at which it seems moderately successful, much like it's success as an interceptor. All it needs to do is to get over a barrel. Would that be a barrel roll? Or a roll out the barrel?