Best Rate of Climb at High Altitudes

[hawk37]

I’m trying to figure a way to experimentally determine the best rate of climb speed at high altitude for a particular jet. Hypothetically, suppose I was to level off at a mach well below the best rate of climb speed (but at a mach where the jet can still accelerate), and then timed each subsequent increase of .01 mach.

Would it be correct to conclude that the least time between these one-hundredths of a mach would give me the mach for best excessive power, and thus the best rate of climb mach?

Or could I only conclude that I have found the mach for most excessive thrust, in which case I’ve found the mach for best angle of climb?

I’m assuming here thrust is constant, no wind, weight, nor temperature change.

I’m thinking this is the easiest way to determine best ROC, other than this manufacturer’s very generalized data? If this were done a few times at the same altitude, one could get a fairly close figure. The alternative, to log rate of climb for various machs over a number of days allows differing air temperatures to affect the results.

All hypothetically speaking.

Hawk

[balsa model]

Assuming that you can losslessly convert this excess thrust to lift by changing attitude (in other words extra induced drag power will balance climb power exactly),

Assuming that changing your attitude (new aerodynamic configuration) will not significantly affect the point of "best excess thrust",

Then:
a*M=f, ('a' best acceleration, 'M' mass, 'f' excess thrust)
f*v=P=roc*M*g, ('v' horizontal speed, 'P' excess power, 'g' Earth acceleration)
Hence:
roc = a*v/g (best rate)
And:
roc/v = a/g (best angle)

Well, it looks like you've found best angle.
You should be looking for best "acceleration * speed" product for best rate.

Just speculating on my lunch time.
If I'm totally wrong than, boy, do I need to be corrected... I can't wait.

[hawk37]

I follow "most" of what you've said. So I'm thinking now then that, as per my example, if I time each .01 mach increase, then the mach which gives me the least time (ie best acceleration, best excess thrust) will be the mach for best gradient climb.

And further, the mach for best rate of climb would be where the product of the acceleration * velocity (best excess power) is the greatest, which thus should be above the best gradient mach as expected.

Thanks Balsa, just please chime in if what I've said is wrong.