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I am aware that the efficiency of the propeller implies a difference in the values of the theoretical power required curve and the corresponding shaft power required, but does that efficiency correction change the abscissae of the respective minima...?

In this diagram (taken from von Mises' 'Fluglehre') the power is specified as 'Zugleistung', i.e. as the theoretical power drag (thrust) x velocity. All usual power required diagrams use that 'theoretical power'. My question is: If we took instead the shaft power, would the airspeed for the minimum of shaft power be the same...?

enter image description here

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  • $\begingroup$ I think you mean "I am aware that the efficiency of the propeller implies a difference in the values of the theoretical power AVAILABLE curve and the corresponding OUTPUT shaft power"? $\endgroup$ – JZYL Oct 11 '19 at 11:43
  • $\begingroup$ No... I mean the power required curve... If you draw the power required curve with the data of the shaft power, you'll get a curve that is different from the one you may construct with a theoretical power that is the product of thrust/drag and airspeed. The reason for the difference is the efficiency of the propeller, that varies with the airspeed. I wonder if the minima of the two curves lie at the same value of airspeed... $\endgroup$ – xxavier Oct 11 '19 at 11:56
  • $\begingroup$ What do you mean minima of the two curves? Do you mean the maximum excess power, or the global maximum/minimum of the power available/power required? $\endgroup$ – JZYL Oct 11 '19 at 12:05
  • $\begingroup$ In other words: at any point of the power required curve you are in unaccelerated, s/l flight. The power you require for that situation is obviously that of the ordinate of the point of the curve where you are. Now, that power implies a value of shaft power that is different. If, with that data for different airspeeds, you build a 'shaft power required curve', would the minima of both curves be the same...? $\endgroup$ – xxavier Oct 11 '19 at 12:06
  • $\begingroup$ It would have to be a large change in efficiency to change the point at which minimum throttle is required. Under steady state conditions (where time and distance are constant), mimimum drag airspeed would be minimum (Power) throttle, even though this is not the maximum power available (efficiency) airspeed. $\endgroup$ – Robert DiGiovanni Nov 12 '19 at 17:36
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The OP is asking about power required, not power available, which, in my mind, would change the answer.

From an aircraft performance perspective, power required is the product of total aircraft drag and inertial speed ($P_R=TV$), applicable only for steady unaccelerated flight. Thus, the minimum of that curve is dependent on profile and vortex drag. In potential flow, it would be where $3C_{D_0}=C_{D_i}$ (see this answer for derivation).

Shaft power required is the power needed to be transferred to the propeller in order to achieve the aforementioned $P_R$. From the actuator disk theory, we know that it must be larger than $P_R$. The shape of this curve would depend also on the propeller itself (e.g. pitch, diameter, airfoil) and its control system (e.g. constant vs variable pitch). Therefore, the minima of the two curves do not have to coincide.

To illustrate this, I've gathered some RC propeller data from APC, which according to its website are generated based on vortex (potential) theory. The drag polar is based on a joke planform designed to accentuate the curvature of the polars:

  • $C_{D_0}=0.02$ (this is way too low for an RC planform)
  • $e=0.85$
  • $W=25lb$
  • $A=8$
  • $S_{ref}=15ft^2$

The figure below shows the power required (P_R) for level flight and the shaft power (torque * rotation speed) needed to sustain the level flight for a few prop configurations.

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  • $\begingroup$ This is good work. Now, to apply this to full scale, even for an electric motor, we will have a huge range of RPM to consider as the 16 x 8 is almost 3 times as long and twice as coarse as the 6 x 4. But what you are showing is these props are more efficient at a speed higher than min sink rate, but the larger, slower one is closer to the power required minimum. Engine efficiency is part of the explanation for Vy. Could you run 16 x 4, 16 x 8, 16 x 10? This would be closer to a variable pitch 25 lb model. Like the graphs! $\endgroup$ – Robert DiGiovanni Oct 13 '19 at 22:55

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