Why still no dive acceleration difference?

[Crumpp]

Okay that is helpful. What Blackberry is wondering, we know the software does these calculations continuously every few milliseconds. In a high speed dive beyond Vmax where TAS is increasing, does the software recognize the degradation of this value? -> Thrust in lbs = 1000lbs. Or maybe it handles it by applying effectiveness factor to this value ->Drag in lbs = 500? Or maybe it ignores it and keeps both opposing forces as static (i.e., "maximum excess force you will have available")?

I can't answer any specific questions about IL2 FM. If it uses standard methods to predict aircraft performance, then it does account for it.

I would be willing to bet it does use standard methods.

Now you know the specific numbers for thrust and drag change. For example here is the P47D22 at Take Off weight from Vs to Vmax:

Thrust available in Pounds
6353.75
6353.75
6353.75
6051.190476
5776.136364
5294.791667
4887.5
4538.392857
4235.833333
3971.09375
3737.5
3630.714286
3529.861111
3434.459459
3344.078947
3258.333333
3176.875
3099.390244
3025.595238
2955.232558
2888.068182
2823.888889
2762.5
2703.723404
2647.395833
2593.367347
2541.5
2491.666667
2443.75
2425.09542

Thrust required in Pounds
1537.476184
1537.476184
1537.476184
1458.574233
1396.29734
1311.942922
1269.557568
1259.393629
1274.840011
1311.284626
1365.430601
1398.36277
1434.870602
1474.747484
1517.813624
1563.911983
1612.9049
1664.671298
1719.104337
1776.109457
1835.602718
1897.509403
1961.762824
2028.303316
2097.07736
2168.036844
2241.138416
2316.342935
2393.614985
2425.09542

[MadBlaster]

yes, from your earlier post on parasitic drag force verse velocity and EAS method.

I suppose you got those numbers from customizing udpgraph tool? I think you must manually configure it somehow, or do calculation off the output file? I don't see those parameters in my version.

[BlackBerry]

We are going to just simplify the drag and thrust values to illustrate the mechanics of solving the problem.

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:
The airplane will move to equilibrium

Entry speed = 300mph = 441fps
Zoom Angle 45 degrees
Vy = 150mph = 220.5fps

Zoom height:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs - 6364lbs = 5864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 220.5^2)/(2 * 20.98ft/s^2) = 3476.18 ft

3476.18 ft * sin 45 = 2458 ft

Crumpp，when airplane zoom 45 degree from 300mph to 150mph, It takes a quite long time: 20 seconds?

In the whole zoom process, Is the acceleration a constant or a variable ? All your formular is of "constant accelerated motion". That's not correct.

In fact, when airplane begin to zoom, the "a"=20.98ft/s^2, however, after a few seconds, as speed drops, the engine thrust increase to 1100lbs, and the air drag will be 450lbs, so the "a" is no long 20.98, "a" will be smaller than 20.98. The acceleration changes during the whole zoom process, so it's a "variable accelerated motion".

I am surprised, your math model is too simple to get correct result.

[Crumpp]

All your formular is of "constant accelerated motion".

This is the expression for the average acceleration:

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

It is all in the formula, blackberry.

Think about it. Do you know what the acceleration would be if just gravity is acting alone on the airplane?

32.2ft/s^2.....

Do you think our acceleration is going to greater or smaller than the acceleration of gravity at the beginning of the dive when we have the most excess thrust?

GREATER THAN

Why is our average acceleration lower than that of gravity???

20.98 ft/s^2 < 32.2ft/s^2......

Because there is an aerodynamic braking force acting on the aircraft that lowers the average acceleration.

Yes the formulation requires constant acceleration but that does not mean it does not accurately describe the motion of the aircraft.

[Crumpp]

Do you mean when the airplane reach 624fps(424mph), the acceleration is zero?

For this airplane under these conditions...YES.

It is called terminal velocity.

[BlackBerry]

Originally Posted by Crumpp

This is the expression for the average acceleration:

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

It is all in the formula, blackberry.

Think about it. Do you know what the acceleration would be if just gravity is acting alone on the airplane?

32.2ft/s^2.....

Do you think our acceleration is going to greater or smaller than the acceleration of gravity at the beginning of the dive when we have the most excess thrust?

GREATER THAN

Why is our average acceleration lower than that of gravity???

20.98 ft/s^2 < 32.2ft/s^2......

Because there is an aerodynamic braking force acting on the aircraft that lowers the average acceleration.

Yes the formulation requires constant acceleration but that does not mean it does not accurately describe the motion of the aircraft.

g=32.2ft/s^2=9.8m/s^2

At the beginning of 45 degree dive, dive acceleration is greater than g due to bigger engine thrust and smaller air braking force, and at the end of dive, acceleration is smaller than g due to smaller engine thrust and bigger air braking force.

But we need to calculate acceleration very accurate. In a 15 seconds long 45 degree dive, if my average acceleration is slightly than yours, for example, is 3ft/s^2 or 0.9m/s^2 more than yours. I'll be 50km/h faster than you in the end. That's a hugh advantage.

Do you remember the 60 degree dive between P47D and fe190G in 1943 summer? At the end of dive, P47D had a much greater speed than fw190G, sth. like 50km/h difference is huge enough.

I have a question for you:
At 1st second, you begin to 45 degree dive,
At 2nd second, your speed increase a little, acceleration is greater than g.
at 3rd second, your speed increase a little more, and acceleration is still greater than g.

.....

at 20th second, you reach the equilibrium point, and acceleration is ZERO. Your speed remains same.
at 21st second, your speed is same as 20th second.
at 22nd second, hour speed is same as 20th second.
......

How about the 19th second? what's your acceleration? It should be smaller than g, but I need more accurate data.

Is the acceleration at 19th second nearly zero or around 0.5*g?

[Crumpp]

I have a question for you:
At 1st second, you begin to 45 degree dive,
At 2nd second, your speed increase a little, acceleration is greater than g.
at 3rd second, your speed increase a little more, and acceleration is still greater than g.

.....

at 20th second, you reach the equilibrium point, and acceleration is ZERO. Your speed remains same.
at 21st second, your speed is same as 20th second.
at 22nd second, hour speed is same as 20th second.
......

How about the 19th second? what's your acceleration? It should be smaller than g, but I need more accurate data.

Is the acceleration at 19th second nearly zero or around 0.5*g?

http://philsci-archive.pitt.edu/1197...y_Solution.pdf

[Crumpp]

At the end of dive, P47D had a much greater speed than fw190G, sth. like 50km/h difference is huge enough.

Look at what the aircraft are placarded for in terms of dive speeds.

The aerodynamic forces will tell you the FW-190A can outdive the P47. Just look at the sea level speeds and power required. At sea level, EAS = TAS and EAS is the speed the airplane always feels.

The P47 generates ~272 THP more than the FW-190A8 to travel ~20mph slower. It takes a lot of power to push that big heavy P47 through the air.

However, the relationship of thrust and drag is not the primary limiting factor in a dive for these airplanes. Dynamic pressure limits and mach limits tend to set the speed limits in WWII fighters.

In the case of the FW-190 vs P 47, the FW-190 is limited to ~466mph at low altitude while the P47 is limited to 500 mph at low altitude.

Those placard limits are not set arbitrarily nor is there any wiggle room or safety factor. A Focke Wulf pilot exceeding 466 mph is taking a huge risk he will not survive the dive. There are plenty of incidents of FW-190 pilots diving straight into the dirt barrier because mach effects made the elevator control ineffective. There are also incidents of the pilot turning the aircraft to confetti by aggressive use of the trim to recover.

In a dive to Vne, the P47 will always gain ~34mph over the Focke Wulf FW-190A.

[BlackBerry]

Originally Posted by Crumpp

http://philsci-archive.pitt.edu/1197...y_Solution.pdf

Come on, Achilles and the Tortoise story is very simple.

Sum of time=1+0.5+0.25+0.125+........ is limited, will never bigger than 2.
of course, Achilles can't over take tortoise WITHIN 2 seconds because tortoise is some distance ahead of Achilles in the beginning.

In ancient time, people didn't know the property of "infinite series", so they were puzzled. They drew the conclusion that Achilles would never overtake tortoise because there are infinite numbers of time series in the sum, We modern people will laugh at them because the sum of infinite numbers is limited to "2".

My story, is totally different, from 1st to 20th seconds, a 45 degree diving a/c acceleration changes from g(usually higher) to 0. During the dive process, a/c moves to equilibium point. The acceleration is always decreasing. If you treat it as a constant acceleration motion, you'll make a mistake.