With a flat rated engine Would a aircraft have the ability to fly at faster indicated air speed as it climbed or would it simply hold the same indicated airspeed ?
It’s speed over the ground would be increasing if it did right ?
Somebody needs to do some basic physics.
Indicated airspeed measures dynamic pressure calibrated for sea level pressure and temperature, and tells us the forces of the airflow over the airplane. q = 1/2*rho*v^2, where rho is fluid density and v is true velocity. In slugs/ft^3 and ft/s, it comes out in lb/ft^2 or psf. If you take an airplane with q= 50psf and 3 sf of effective drag area (product of Cd and S), that is 150 pounds drag. It matters not at all how you get q=50 psd, you get the same drag if your airspeed indicator is reading the same.
Power is force applied through a distance in time we are not talking a theoretical distance, real distance swam through the air. Power = F*dx/dt = F*v.
Applying some numbers. At sea level, rho is 0.002378, and v to get q=50 psd is 205 fps or 122 knots true. Power is thus 150*205=30750 ft-lb/s or 56 hp.
At 10,000 feet, rho is 0.0017555, and v to get q =50 psf is 239 fps or 142 knots true. Same 150 pounds total drag. Power is 150*239=35850 ft-lb/s or 65 hp.
I put to you, through some really easy physics that it takes MORE power to fly same airplane to same indicated airspeed at higher altitude. Your flat rated engine will not take you to the same indicated airspeed high as it did at sea level - it does not have enough power...
Let's answer the "same power" question at two altitudes. First off we have to make some assumptions. Is the flat plate equivalent area the same as we slow down? Usually it drops some as we fly at lower q, but for the sake of keeping it simple here, let's say it stays the same.
Power = F*v = 1/2*rho*v^2*A*v = 1/2*rho*A*v^3
So for same power 1/2*rho1*v1^3*A = 1/2*rho2*v2^3*A, the 1/2's and A's are the same, so we reduce things a bit:
rho1*v1^3 = rho2*v2^3, and solving for v2
v2 = cube root(rho1/rho2*v1^3)
Given the sea level case, with 0.002378 spcf, 205 fps, 122 knots true, and 56 hp, while the 10,000 foot case of 0.0017555 spcf, you will go 227 fps, 135 knots true, and 110 knots indicated.
So, the flat rated engine will indicate lower as you go up, but will be going faster through the air as it goes up. Just not a huge amount and all airplanes get this same effect.
If you are wondering about how much a naturally aspirated engine loses with altitude, its power basically drops linearly with atmospheric density.
Please play with the inputs yourself and answer your own "what if" questions, here is the atmosphere
A Table of the Standard Atmosphere to 65,000 Feet in US units, and a little Excel time will let you work out any question you want.
Billski