You are beginning to understand me.

"X" vs "Y" ,Everything ELSE being euqal, propellers are different: good prop vs bad prop.

During TAS 0.2-0.6 Mach, good prop=bad prop=85%

During TAS 0.6 -0.8 Mach, good prop> bad prop; ie 80%>60%.

"Y" has no more gravity for help because both dive in same angle=45 degree, and both weight=5 tons.

"Y" has no reinforcement.

the example works better for me if you assume the 'sudden' drop in efficiency for the bad prop happens at the margin of the csp peak envelope at/near Vmax/level. And to make it even more emphasis, the efficiency drops from .85 peak to 0 no thrust when 1 more kph is added above Vmax/level in 1 second, as soon as you enter the 45 dive. also, drag force just kicked you back into the peak envelope. okay, yes the bad prop now has excess negative thrust. because in that 1 second you went from max thrust to no thrust from the prop and drag force pushed you back and not enough time has elapsed for excess thrust from 45 weight vector to overcome the momentary loss from the prop. I'm probably missing something.


Now, we know there are no 'sudden' drops in these curves. they have slopes and they are a functions of TAS, RPM, reduction ratio, blade diameters...etc. this example, rigged.

Good prop and bad prop share same diameter, same rpm, same advance ratio@same TAS. Their DIFFERENCE IS THE SHAPE OF AIRFOIL SECTION.

TAS Mach / good CSP / bad CSP
0.6 85% 85%
0.61 84% 83%
0.62 83% 80%
0.63 82% 77%
0.64 82% 73%
0.65 81% 70%
0.66 81% 67%
0.67 81% 64%
0.68 80% 60%
0.69 80% 56%
0.70 80% 52%

//////////////////////

Your TempestMKV equipped with bad propeller, my Tempest with good propeller.

If I drag you into a high speed dive, around 0.7 Mach TAS for 40 seconds, what is your energy loses?

To simulate this sharply efficiency drops for bad propeller, you can simply use il2 4.11m, quick mission, spawn @ 5000 m with Tempest mkv, you intently decrease your throttle to 60% during 0.6-0.7 Mach 45 degree dive and 45 degree zoom.

You will feel the energy loss.

putting the weight vector/dive angle to the side for a bit, I think what is missing here in the discussion is the drag vector. Now, beyond peak/vmax. what is happening is the drag vector is slowly overwhelming the thrust of the propeller as it gets more and more inefficient until efficiency is at zero and advance ratio reaches some high number at some given high TAS. So, it is negative excess thrust between those two vectors because drag force is becoming greater than the prop thrust vector (still ignoring the weight vector for now). So, the question is how is the interplay between these two vectors modeled in IL-2. Maybe it is already accounted for in the modeling of the drag force? Like an effectiveness factor or something that is attached to the drag coefficient as the thrust begins to degrade beyond Vmax? I really have no idea, only speculate. :)


edit:
of course, the only way to get beyond vmax is to dive, and its the sum of all forces...etc. so this interplay between opposing thrust and drag vectors gets masked over by weight/dive vector. am i on the right track?

Airplanes certainly do reach their equilibrium point in a dive.

However the most common restriction to dive performance is dynamic pressure limits <flutter> and mach limits.

Completely irrelevant though as the equilibrium point estabilishes the rate of aceleration in the dive.

You can debate it all day long but it does not change the fact it is how performance is predicted.

The excess force is what drives the aircraft to equilibrium. It is all accounted for in the formulation.

The interplay is covered in the derivative.

You start out with the maximum force which is the moment in transitioning that a component of weight has shifted to thrust and the propeller thrust is still at level flight velocity. That is the maximum excess force you will have available.

Let's look at the rectilinear motion equations and solve both a zoom climb problem and a dive problem using the same airplane at the same entry speeds.

We will end our zoom at Vy or best rate of climb speed and end our dive at the equilibrium point.

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

Now let's dive under the same conditions:

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:

Entry speed 300mph = 441fps
Angle of Dive 45 degrees

We need a ballpark of our equilibrium speed. A quick method is to use the relationship of Parasitic drag. It is the major drag component at high speed. More detailed analysis will give better results but this is accurate within 10%. 10% is acceptable for climb/dive performance.

441(SQRT 1000/500) = 624fps

At 624 fps, the acceleration about the CG will be zero.

Dive:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs + 6364lbs = 6864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 6864lb/279.5lb-s^2/ft
a = 24.6 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 624^2)/(2 * 24.6ft/s^2) = -3961ft (negative is a vector direction)

3961ft * sin 45 = 2801 ft of altitude lost!!

There are very few airplanes with a "bad propeller". It does happen but is caught very quickly as nothing will line up between predictions and the realized.

Including dive performance....which is how they caught it.

That is why Hawker initiated the study.

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

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.

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.