Aerodynamics question

[moot]

Suppose you've got a speed/altitude chart with lines for two of a plane's possible weights..

That's a plane at 9.5 and 11.3 tons, with two engines running at 1.3 ata and 2500rpm for 1580hp...

..and one more line at unknown weight and a different throttle/rpm setting than the other two lines:

That's an assumed 2700rpm at some unknown manifold pressure for 1750hp at sea level, and 1850hp at ~7kft.  

Could you extrapolate from these points some ballpark lines for different weights with the maximum power engine settings?

[Stoney]

At first glance, I was going to say yes, but then I noticed the relationship between the top two blue points.  The top blue point at 26,000 feet doesn't maintain the same relationship to the other two points, at that altitude, as the two plotted points do, at least to my eyeball.  It looks as if the speed decreases more rapidly as a result of altitude than as a result of the decreased weight.

This is probably more of a math question than an aerodynamic one.  There may be a formula out there you could use, but I'm at a loss.  Is there any way to plot more points along the unknown curve?

[moot]

I'm pretty sure these are all the data points available.  Just to be sure I communicated this clearly:
The red points are 1.3 ata and 2500rpm, on the 9.5ton line, whereas...
The blue points are without a given weight. They're given only as "top speed" quotes.  It's my assumption that they're with max. engine power settings, that is, 1750hp @ SL and 1850hp @ 7kft... 1.3ata / 2500rpm, and 1.x ata / 2700rpm respectively.
 
Maybe there's a variance in power output (manifold pressure?) for the blue line, whereas the lower output settings (1.3 ata, 2500rpm) could be kept constant thru the whole altitude range?
I'll keep looking for some way to figure out either the missing ata values, or the weight of the plane they used for the max power trial.

[Stoney]

Perhaps I wasn't clear.  If you drew a curve using extrapolated points, it appears that the blue curve would not match the relationships of the solid and dashed curves once you got above the critical altitude.  That being said, I may be wrong--there may be a math application here that I don't know.

From an aerodynamic standpoint, if weight is fixed, the only thing that changes with altitude, generally speaking, is dynamic pressure and density altitude.  So, for example, if the first stage of the supercharger peaks at 3500 meters and the second stage peaks at 5250 meters, it would peak at the higher RPM and MP settings at the same altitudes.  That's why the first two red curves hold a similar relationship all the way up to their ceilings.  Notice the part that I highlighted--the slopes of the red lines are close.  The slope of the blue line is much flatter than the two red slopes.  I would be careful about extrapolating points due to this.  If the slope of the blue line was similar to the red lines, I would feel more comfortable with a result from extrapolation.



Perhaps you could go ahead and extrapolate some points and see what the curve would look like?  Is there any way to test the points once you've plotted them?

[moot]

No Stoney, I don't have any aerodynamics know-how.   I understood you right the first time.  What I was trying to suggest as a possible reason for the discrepency between the 9.5/11.3-ton lines is that they may have run those two trials (in red) with the manifold pressure and RPM set to fixed values, 1.3 ata and 2500rpm. That's what the title for the chart is.
Click for the full size version:


In my uneducated eyes, that could account for the difference in slopes between the red and blue lines: at full throttle the engine output variance couldn't be helped (just like we see in AH, firewalled manifold pressure going back and forth with altitude), whereas at less than full throttle they could always keep up with that altitude-induced variance in manifold pressure.

My next guess would be that, given those speeds for two specified weights, and assuming that they were made with constant MAN/RPM settings, could we not deduce the drag coefficient?  Isn't that all we'd need to then deduce the weight of the plane used in the max speed trial, in blue? 
Components of a drag coeff are supposed to be (from wikipedia):
Fd is the drag force, which is by definition the force component in the direction of the flow velocity. - Meaning aircraft thrust?  Couldn't we consider this constant between any pair of points on the 9.5t and 11.3t lines?
ρ is the mass density of the fluid - This ought to be easy enough to figure out, given AH's idealized air model.
v is the speed of the object relative to the fluid - We know this one, it's given on the graphs.
A is the reference area. - A constant since the plane is the same in all trials.  Would it be hard to deduce this?  Would there be any significant funny effects varying with speed, where the plane has different effective areas depending on speed or altitude?

There's no way to test extrapolated points that I know of, because this is for an Me410, which we don't have in AH.  I'm going to keep reading the books I just got for more clues..


edit- Actually, there should be no way they could have run this test at constant weight.  It had to have been the weight at the start of the trial run, if not weight on take-off.  Maybe  the question I'm asking can't be answered with so little info.. Without knowing the engine's GPH....  Or maybe they knew enough aerodynamics to run the trials only at the cusps of the graphs, and set the plane up to weigh exacly 9.5 or 11.3 tons at each of those 4 altitudes.  Then connected the dots.

[Stoney]

What I was trying to suggest as a possible reason for the discrepency between the 9.5/11.3-ton lines is that they may have run those two trials (in red) with the manifold pressure and RPM set to fixed values, 1.3 ata and 2500rpm. That's what the title for the chart is.

Ok, I'm not very good with German so I didn't know that.  Then obviously, we're looking at simply the effects of weight on the two plotted curves.

the engine output variance couldn't be helped (just like we see in AH, firewalled manifold pressure going back and forth with altitude), whereas at less than full throttle they could always keep up with that altitude-induced variance in manifold pressure.

No, what you're seeing with that zig-zag path of the curves is the effects of the supercharger gearing, regardless of whether or not they're using full throttle.  Density altitude, assuming a standard lapse rate, would give you a smooth curve, akin to what you see on P-47 power curves.  multi-staged supercharged planes almost always have a zig-zag plot (as long as the supercharger is not used to "normalize" manifold pressure, i.e. maintain sea level MP and allow no more than sea level MP up to a specific altitude) as the gears only hit their peak efficiency at a single altitude.  On a supercharged aircraft, if you took off with the throttle lever set in a specific position and left it there as you climbed, even if you took off with a reduced power setting, your plot would have the zig-zags as a result of the supercharger's varying degrees of efficiency, especially as it changed gears.

My next guess would be that, given those speeds for two specified weights, and assuming that they were made with constant MAN/RPM settings, could we not deduce the drag coefficient?  Isn't that all we'd need to then deduce the weight of the plane used in the max speed trial, in blue? 
Components of a drag coeff are supposed to be (from wikipedia):
Fd is the drag force, which is by definition the force component in the direction of the flow velocity. - Meaning aircraft thrust?  Couldn't we consider this constant between any pair of points on the 9.5t and 11.3t lines?

No, because at higher weights, induced drag is higher for the heavier plane.  In matter of fact, induced drag increases exponentially (^4) with increases in altitude, due to decreased lift as a result of lower dynamic pressure.  The same plane at a higher weight has to fly at a higher angle of attack to create the lift required, and therefore will create more induced drag (more lift generated) as well as increased profile drag on the wing.  So, in sum, the same aircraft at a higher weight will have a different, higher effective coefficient of drag.

ρ is the mass density of the fluid - This ought to be easy enough to figure out, given AH's idealized air model.
v is the speed of the object relative to the fluid - We know this one, it's given on the graphs.
A is the reference area. - A constant since the plane is the same in all trials.  Would it be hard to deduce this?  Would there be any significant funny effects varying with speed, where the plane has different effective areas depending on speed or altitude?

You're correct on these three.  The effects of density are best expressed as dynamic pressure, as I alluded to earlier, and yes, with AH idealized atmosphere, it would be easily computed for increases in altitude.  Reference area would also be constant, as well as pressure drag from the wetted surfaces of the aircraft.  Unfortunately, as speed increases, parasitic drag increases, so there's another variable thrown in for you. 

It had to have been the weight at the start of the trial run, if not weight on take-off.

Yes, typically when these types of tests are done, take off weight is the measurement, since fuel is consumed in the climb.

[bozon]

OK, lets look at it this way (feel free to correct my assumptions, I am not an expert):
Lets assume that for small angle of attack (AOA), both induced drag and lift are proportional to the AOA in the same way (i.e. changing the AOA affects both by the same factor. Note that we define AOA=0 for the angle of zero lift), the airspeed squared and the density.
The lift is equal to the weight in level flight.
Therefore, we can replace the density, square of the airspeed and angle of attack in the induced drag equation with lift or weight -> we get that the induced drag is proportional to the weight (times some constants that represent aerodynamic efficiency of lift per drag): drag_i ~ M
It seems strange that this is constant, but we are assuming high speeds, reasonable weights and therefore very small variations in the AOA.

thrust = drag_parasit + drag_induced

thrust is a function of alt. so lets keep alt constant. parasitic drag is proportional to v^2 -> drag_parasit~v^2
Therefore, for the two curves at a given alt you have two airspeeds: v1 and v2.
Lets make a mistake (* explained below) and subtract the above expression for curve 1 from the same expression for curve 2 at a given alt. We get:
0 = A*(V2^2-V1^2) + B*(M2-M1)
or:
V2^2-V1^2 = C*(M1-M2)

Where A,B,C are just combinations of constants. So you get C by putting the values of the masses and airspeeds of the two points. Now that you know C you can assume a new mass M3 and find V3 from the same equation relative to, lets say (M1, V1) values.

* the mistake is that thrust is not constant with speed. Propellers are designed to be efficient at a given speed range, usually trading low speed climb/acceleration rates for max speed, or vice verse. A large diameter propeller may have restriction on RPM at high speeds due to the tips breaking the sound speed. How significant is this to the above case? I don't know. If it is not extreme in the relevant speeds range than this could be a valid "ball park" approximation (if the other assumptions about AOA in drag and lift are valid...).

example:
at 5500 m alt we have:
V1=570 M1=11.3
V2=590 M2=9.5
this gives:
C=(590^2-570^2)/(11.3-9.5) = 1.29E+4

so, for 12 tons:
V3^2=C*(11.3-12)+570^2
V3 = 562 km/h

and there you have it. Most likely wrong, assumption assumption... but still...  just note that C is different for any given alt, but I suspect it should go like 1/ro the density. In principle, measuring C at different alts (thus different speeds in the chart) and accounting for ro would give you the effective thrust and some combination of the lift and drag coefficients.

[hitech]

the mistake is that thrust is not constant with speed. Propellers are designed to be efficient at a given speed range, usually trading low speed climb/acceleration rates for max speed, or vice verse.

This statement is a little miss leading, and I am not sure if you are considering the following.

To begin with Thrust = HP/speed. This is the basic equation of thrust for constant speed propellers.

As you said Larger vs smaller and faster spinning vs slower spinning, and Paddle blade vs non paddle blade effects things but not this equation.

2 things happen at both end of the speed ranges. At the slow end of the speed range, I.E. standing still and very slow speeds, if the propeller is very small it can not create any more AOA on the blades with out stalling the blade, hence Prop efficiencies goes down.

At the higher end of the speed range as you stated tip speed can aprouch supersonic and hence create more drag on the prop, hence less efficiency.

A 100% eff prop would maintain the equation above. But most props a swag number of 82% eff can be used for normal ranges. So

Thrust = HP / speed * PropEff.

The PropEff is what is effected by the specific prop characteristics.  As an example when some of the US planes went to a paddle blade prop it lowered the PropEff a little at higher speeds, but it greatly increased the PropEff in the best climb speed range.

HiTech

[gripen]

I used a bit different approach based on couple assumptions:

First assumed that the e is 0,75 (Oswald's efficiency) and exhaust thrust 90kp per engine. Then matched the speeds at sea level and 1580ps per engine for 9,5tn and 11,3tn by changing the propeller efficiency and 80% gave 489km/h for 9,5tn and 483km/h for 11,3tn.

Then using the same values but 1750ps and 100kp exhaust thrust per engine I calculated that it would have gone at sea level 509km/h at 9,5tn and 503km/h at 11,3tn.