Author Topic: Oh my god the pony is awful! (Read 8468 times)

Widewing · « **Reply #135 on:** February 26, 2008, 06:00:02 PM »

Quote

Originally posted by Tilt
So if I have a Cx of 0.0228 (trimmed) and 0.0273 (untrimmed)

an all up weight of 7115 lbs
and a max take off power of 1850 hp

(36.6lbs/ft^2 wing loading)

How would you compare them to the P51 and the Tempest under study above.

What is the wing area?

My regards,

Widewing

Captain Virgil Hilts · « **Reply #136 on:** February 26, 2008, 06:15:24 PM »

Quote

Originally posted by SgtPappy
Not to shoot down your statements, Widewing, but if they purposely limited the RPM which lowered the maximum power output, would that not mean there was a reason for the lowered RPM; i.e. they lowered the RPM (and consequently, power) for a reason.

Now did the pilots flying these P-38L's commonly manage 1,700 + hp using the Allison F-30's? If not, that would be the reason as to why we do not have the max. power output. Whether the F-30's had the ability and potential to produce that power is irrelevant... it's whether it was used commonly or not... I think. I could be wrong. I'm not sure how HTC controls how they add features to planes.

YES, 70"+ MAP @ 3200 RPM WAS commonly used. NO, it was not an accepted power setting in MOST documents. That is why it (P-38L) does not have that setting in AH. Lockheed reps did turn the planes up in the field, and they also showed mechanics and crew chiefs how to do it.

Captain Virgil Hilts · « **Reply #137 on:** February 26, 2008, 06:18:20 PM »

Quote

Originally posted by Stoney
Obviously, those differences would be significant. I assume all of those are Sea Level comparisons. Given that the P-38 is turbo-supercharged, I'm assuming the Allisons made 1725 up to altitude (a la P-47). I'm curious as to standard day altitude comparisons that could create an advantage for the sleeker Pony.

I'm having some trouble getting XFoil to accept the coordinates for both the Tempest and P-51 airfoils as listed on the UIUC website. I'll play around with it to see if I can get some results.

Most P-38's from the mid to late G on (and probably earlier) could maintain full boost on up to around 29,000 to 30,000 feet. The limit, same as the P-47, was turbine RPM in the turbochargers. There's an armored ring around the turbocharger for a reason. If you exceed the RPM limit, there's a serious risk of turbine explosion.

Widewing · « **Reply #138 on:** February 26, 2008, 06:21:15 PM »

Quote

Originally posted by SgtPappy
Not to shoot down your statements, Widewing, but if they purposely limited the RPM which lowered the maximum power output, would that not mean there was a reason for the lowered RPM; i.e. they lowered the RPM (and consequently, power) for a reason.

There was a reason. Standardization. Plus, previous issues with the F-17s.

From what I have learned talking with pilots and crew chiefs, the P-38Ls were almost universally adjusted for max RPM after the aircraft was received by the various Groups.

By the way, I have no issue with HTC using the standardized 1,600 hp rating. That's what the AAF decided to rate them at. Of course, both Allison and Lockheed did not.

In late September of 1945, a memo from 5th AF HQ was circulated through the 49th FG directing fighter squadrons to adjust the prop governors on the P-38s to peace time settings. According to Bill Pascalis, the 49th's historian, this meant resetting the engines to 3,000 RPM.

My regards,

Widewing

Lumpy · « **Reply #139 on:** February 26, 2008, 06:30:47 PM »

If you accept the 1600 hp rating in-game then you should also use 1600 hp in your calculations. Using boost settings that are not available in the game is not relevant if you're trying to argue that the flight model is wrong. It could even be considered misleading.

Tilt · « **Reply #140 on:** February 26, 2008, 06:41:48 PM »

Quote

Originally posted by Widewing
What is the wing area?

My regards,

Widewing

So if I have a Cx of 0.0228 (trimmed) and 0.0273 (untrimmed)

an all up weight of 7115 lbs
and a max take off power of 1850 hp

(36.6lbs/ft^2 wing loading.............. = 194.4 ft^2 @ NACA root>tip 23016>23010)

How would you compare them to the P51 and the Tempest under study above.

Widewing · « **Reply #141 on:** February 26, 2008, 07:08:45 PM »

Quote

Originally posted by Tilt
So if I have a Cx of 0.0228 (trimmed) and 0.0273 (untrimmed)

an all up weight of 7115 lbs
and a max take off power of 1850 hp

(36.6lbs/ft^2 wing loading.............. = 194.4 ft^2 @ NACA root>tip 23016>23010)

How would you compare them to the P51 and the Tempest under study above.

One still needs to know what the total drag is, and that will vary with speed and altitude due to the changes in dynamic pressure. Do you have access to wind tunnel data showing the drag at various speeds?

Perhaps someone with more familiarity can calculate the drag based upon the
flat plate area of 4.4574 sq/ft.

My regards,

Widewing

Stoney · « **Reply #142 on:** February 26, 2008, 07:48:02 PM »

Quote

Originally posted by Tilt
So if I have a Cx of 0.0228 (trimmed) and 0.0273 (untrimmed)

an all up weight of 7115 lbs
and a max take off power of 1850 hp

(36.6lbs/ft^2 wing loading.............. = 194.4 ft^2 @ NACA root>tip 23016>23010)

How would you compare them to the P51 and the Tempest under study above.

What aircraft are these numbers for? Certainly not the Tempest?

Widewing · « **Reply #143 on:** February 26, 2008, 08:06:49 PM »

Quote

Originally posted by Stoney
What aircraft are these numbers for? Certainly not the Tempest?

Looks like the La-7 to me.

My regards,

Widewing

Captain Virgil Hilts · « **Reply #144 on:** February 26, 2008, 08:19:24 PM »

Quote

Originally posted by Lumpy
If you accept the 1600 hp rating in-game then you should also use 1600 hp in your calculations. Using boost settings that are not available in the game is not relevant if you're trying to argue that the flight model is wrong. It could even be considered misleading.

He did.

Lumpy · « **Reply #145 on:** February 26, 2008, 09:06:49 PM »

He did? I don't know what middle-school math you guys use but to me 2 times 1600 does not equal 3450.

Quote

Originally posted by Widewing
I'll answer the last question first as I have this handy.

Root Chord @ C/L is 103.99"
Wing Tip Chord at Station 215 (where the wing tip cap is screwed on) is 50.0".

As to the the P-38/P-51 comparison, there's a few things your source omits, if you will..

They have the CDo at .0163. This is for a P-51B/C and it reflects no stores pylons and taped over gun ports. For the P-51D it should be .0176 with pylons and untaped gun ports.

Also to be considered is that the hp rating for the P-38 reflects MIL power, not WEP. It gets tricky because the AAF rated the V-1710-89/91 engines at 1,600 hp in WEP, and applied the same rating to the V-1710-111/113 of the L model. However, the L model was delivered with a factory rating of 1,725 hp. Tony LeVier and Kelly Johnson have both stated that the L models were delivered and accepted by the AAF with the full 1,725 hp available in WEP. However, it was merely an adjustment to the prop governors to limit RPM to 3,000 and 1,600 hp. The full 1,725 hp occurred at 3,200 RPM (according to LeVier). So, the AAF could have readjusted the RPM during assembly and rigging at the Burtonwood or Langford Lodge depots. Of course, the Lockheed and Allison reps immediately reset the RPM when the aircraft were ferried to the combat units (according to several P-38 crew chiefs).

As to the P-51, we have the same issue in that the MIL rating appears in your source rather than the 1,720 hp WEP rating.

If we do the math, we find the following using full internal fuel weights:

P-38J: 375 x .8 x 3,200 / 150 mph = 6400 lb thrust
So, 6,400 lb thrust - 1676 lb drag / (16,480/32.2) = 9.23 ft/sec/sec

P-38L: 375 x .8 x 3,450 / 150 mph = 6,900 lb thrust
So, 6900 lb thrust - 1676 lb drag / (16880/32.2) = 9.97 ft/sec/sec

P-51D: 375 x .8 x 1,720 / 150 mph = 3,440 lb thrust
so, 3440 lb thrust - 845 lb drag / (10,208/32.2) = 8.18 ft/sec/sec

At 250 mph:

P-38J: 4.22 ft/sec/sec
P-38L: 4.70 ft/sec/sec (4.13 ft/sec/sec for 1600 hp rating)
P-51D: 3.85 ft/sec/sec

Much depends upon weight. Early P-38Js didn't have leading edge tanks, so you could subtract the weight of 110 gallons, or 660 lb.

My regards,

Widewing

Stoney · « **Reply #146 on:** February 26, 2008, 10:49:52 PM »

Quote

Originally posted by Lumpy
He did? I don't know what middle-school math you guys use but to me 2 times 1600 does not equal 3450.

No, but 1725X2 does...

Widewing · « **Reply #147 on:** February 26, 2008, 11:00:22 PM »

Quote

Originally posted by Lumpy
He did? I don't know what middle-school math you guys use but to me 2 times 1600 does not equal 3450.

You missed this...

P-38J: 4.22 ft/sec/sec
P-38L: 4.70 ft/sec/sec (4.13 ft/sec/sec for 1600 hp rating)
P-51D: 3.85 ft/sec/sec

My regards,

Widewing

SgtPappy · « **Reply #148 on:** February 26, 2008, 11:05:31 PM »

So they did commonly use the rating, eh?

Well then. I guess it's HTC's call whether or not to use the rating or not. Is there any flight test data that shows this rating?

Tilt · « **Reply #149 on:** February 27, 2008, 03:02:00 AM »

Quote

Originally posted by Widewing
One still needs to know what the total drag is, and that will vary with speed and altitude due to the changes in dynamic pressure. Do you have access to wind tunnel data showing the drag at various speeds?

Perhaps someone with more familiarity can calculate the drag based upon the
flat plate area of 4.4574 sq/ft.

My regards,

Widewing

It is an La-7 and the data is from wind tunnel tests.

However it was to show Cx and Cy against attitude.(not altitude) and two levels of "trim" caused by a change to the fuselage. The lower level is the standard fully feathered setting.

All tests were at approx 40.5m/sec (89mph?) at 756 mm mercury (sea level'ish) at 26 deg C. The only graphs showing variance in air speed are showing Cx at fixed attitudes and then the line is flat regardless of speed.

Cy at 0 degrees attitude was approx .005 but it changes from -.1 at -2 degrees to +. 2 at + 2 degrees with hardly any change in Cx at all.

in the formulae attached showing various calculations Cx=f(V)

a figure for Fa is referred to frequently

Another brick wall I guess. thanks anyway.