[Holtzauge]

Originally Posted by drgondog

Let's take one more pass at this.
Assume max TO power, brakes on, zero velocity. I say that is the condition of Max Thrust Available for the system at that altitude.

The system does not accelerate, no drag forces are experienced and the force retarding the thrust is the friction on the wheels with the brakes creating the torque to keep the wheels from rotating.

Release brakes, Same Max Thrust Available.

The Max Thrust available exceeds the thrust required and the system accelerates until the speed is sufficient to create enough lift to overcome weight. Same Max Thrust but the a/c is airborne and continues to accelerate.

Drag forces increase as the system accelerates until the drag forces equal the Max Thrust Available... at that point the system has reached maximum velocity for that system with that Max Thrust available. It is in equilibrium.

At no time did Max Thrust available change from V=0 through V=Vmax for that weight condition.

Add 500 pounds of fuel.

Everything is the same for this System B except that induced drag increases throughout the profile in comparison to the lighter weight system

MAX THRUST available for the System B is the same as it was in the lighter weight system, but the aircraft will achieve equilibrium at a lower speed because the Thrust required to attain an equal velocity of the lighter weight system is insufficient to overcome the increased induced drag due to the heavier weight system.

Summary

Max Thrust for System A = Max Thrust for Sytem B
Max Weight for System A < Max Weight for System B
Max Velocity for System A > Max Velocity for System B
Max Thrust for System A still = Max Thrust for System B

Think of it another way

System A pops a drogue chute at max speed and Max Thrust
Drag increases dramatically, speed drops dramatically, induced drag drops dramatically, parasite drag increases astronomically,

System A achieves equilibrium at lower speed because the Max Thrust Avaialable is lower than Thrust required to maintain the higher speed.

System A makes no changes to internal weight, makes no changes to power settings or altitude, but slows down dramatically...

So, Harri - by your use of the conversion of Jet Thrust to THP conversion equations do you believe that Thrust just increased in the example above by virtue of the reduction in velocity as described in System A above?

The answer is no.

The Max Thrust Available did not change from zero Velocity to Max velocity, nor did the weight of the system change. Only the parasite drag increased and induced drag decreased causing the System to decelerate to a lower velocity. No Thrust change, no Hp change

That equation is all about converting the thrust of a jet to an estimate of work being performed on a system by a propeller/piston engine system, and with modification, to a turbo prop.

I don't have the Hamilton Standard 'Red Book', but it probably has something similar to page 109 of the Blue Book - THRUST-HORSEPOWER CONVERSION in the GAS TURBINE section.

Ask yourself how real Force increases with a reduction in velocity. If you looked at the two states of System A, the first at rest with brakes applied and the second at Vmax - both with same HP/propeller combo/rpm, by your anology the Thrust would be infinite when the system is not moving?

Drgondog, You seem to have problems understanding the propeller thrust formula Harri is using and it's limitations. In the context that has been discussed here (max speed) it is most certainly valid. However, when the plane just starts moving at very low speeds as in your example it is NOT valid. Since you claim to have an Ms in aeronautical engineereing I'm frankly surprised you do not know this and imply the formula is wrong since the thrust would be infinite at zero speed. Look up any old NACA report dealing with propeller efficiency and you will see that the propeller efficiency goes down to zero at zero speed. So we end up multiplying something approaching zero with something approaching infinity. However, at any speed that is interesting for performance analysis, e.g climb, turn and speed performance the formula is perfectly valid.

All the talk in this thread about the meaning and definition of "thrust horsepower" is as meaningful as discussing how many angels can dance on the tip of a pin. You can use different methods to arrive at a correct answer and I would certainly like to be shown why it is not possible to do the analysis according to the method provided by Harri and why one MUST use "thrust horsepower" or else is doomed to failure.

Using the thrust formula Harri posted above in the analysis has a clear advantage over the power method: For a WW2 figther, there is a significant part of the thrust (especially at high speed) that is derived from exhaust thrust. The exhaust thrust is close to constant with speed. It would be interesting to hear how you account for this in the "thrust horsepower" model.