Torque vs RPM
Posted: Tue Jan 22, 2019 8:43 pm
Hi,
I would like to address my observations on torque change when RPM rises or decreases.
I've read the previous topic where this was mentioned (viewtopic.php?p=82867#p82867) but decided to make a new one specific to that single issue. I hope that's ok
Basically - DHC-3 is powered with a free-turbine PT6A, so the gas generator runs independently of prop RPM (no physical connection between the two). Power lever will, therefore, set gas generator power, resulting in the amount of air blown into the free turbine that is connected to prop. Power output on the prop shaft should be similar to the power output of the gas generator. For the sake of the discussion, we can ignore the energy lost in between.
A little more theory: the power is (basically) torque multiplied by RPM. The formula for power would be Power = Torque x RPM. At this point, we it's unitless. Horse Power formula looks like this: Horse Power = Torque (foot pounds) X RPM ÷ 5252. We could calculate PSI to foot-pounds in Turbo Otter, but it's beside the point, and we don't need to calculate the actual power at any point. We just need to see the relation, which is derived from this equation: Power = Torque x RPM.
Leaving specific units aside we can calculate that power settings of 50 PSI / 2000 RPM should produce 100,000 abstract units of power. As has been mentioned earlier - changing RPM does not change the power (free turbine). So when RPM decreases the change in torque should keep the result the same. For example: At 1800 RPM the torque gauge should show 55.56 PSI. It is not so in Turbine Otter.
In my test flight (ISA, approx. 150 kias, 2000 ft, straight and level) I got the following results:
- 2200 RPM, 48 or 49 PSI
- 2000 RPM, 50 PSI
- 1800 RPM, 52 PSI
- 1600 RPM, 54 PSI
That means that moving prop lever from 2000 to 2200 increases power by 10,6%. And there is a corresponding change in speed.
There shouldn't be. No matter what RPM (within the governor range) - power output should be the same and the speed should be the same. Engine wear will differ. Noise will differ. But not the power (nor speed).
If my explanation is not detailed enough I would recommend this article:
http://blog.covingtonaircraft.com/2015/ ... -vs-power/
You can also see the thing I'm talking about in steveo1kinevo's video:
https://youtu.be/GZv-5RBdugM?t=823
Look at how he changes power settings for the cruise. He flies at following power settings: around 86 % / 2000 RPM. Then he reduces RPM to 1900 and gets around 90 %. Do the math - you will get 172000 from both 86*2000 and 90*1900. Actually, you will get 172K and 171K but this difference is far below the accuracy of me reading his gauges.

PS
Let me say that apart of this issue your DHC-3 is a joy to fly and I really love the work you put into correct simulation of this aircraft and it's engine!
I would like to address my observations on torque change when RPM rises or decreases.
I've read the previous topic where this was mentioned (viewtopic.php?p=82867#p82867) but decided to make a new one specific to that single issue. I hope that's ok

This change seems incorrect.Taguilo wrote:There is indeed an increase in torque with prop reduction, between 2.5 - 3.0 psi between 2200 and 2000 RPM. Full governor range is ~ 8 psi.
Basically - DHC-3 is powered with a free-turbine PT6A, so the gas generator runs independently of prop RPM (no physical connection between the two). Power lever will, therefore, set gas generator power, resulting in the amount of air blown into the free turbine that is connected to prop. Power output on the prop shaft should be similar to the power output of the gas generator. For the sake of the discussion, we can ignore the energy lost in between.
A little more theory: the power is (basically) torque multiplied by RPM. The formula for power would be Power = Torque x RPM. At this point, we it's unitless. Horse Power formula looks like this: Horse Power = Torque (foot pounds) X RPM ÷ 5252. We could calculate PSI to foot-pounds in Turbo Otter, but it's beside the point, and we don't need to calculate the actual power at any point. We just need to see the relation, which is derived from this equation: Power = Torque x RPM.
Leaving specific units aside we can calculate that power settings of 50 PSI / 2000 RPM should produce 100,000 abstract units of power. As has been mentioned earlier - changing RPM does not change the power (free turbine). So when RPM decreases the change in torque should keep the result the same. For example: At 1800 RPM the torque gauge should show 55.56 PSI. It is not so in Turbine Otter.
In my test flight (ISA, approx. 150 kias, 2000 ft, straight and level) I got the following results:
- 2200 RPM, 48 or 49 PSI
- 2000 RPM, 50 PSI
- 1800 RPM, 52 PSI
- 1600 RPM, 54 PSI
That means that moving prop lever from 2000 to 2200 increases power by 10,6%. And there is a corresponding change in speed.
There shouldn't be. No matter what RPM (within the governor range) - power output should be the same and the speed should be the same. Engine wear will differ. Noise will differ. But not the power (nor speed).
If my explanation is not detailed enough I would recommend this article:
http://blog.covingtonaircraft.com/2015/ ... -vs-power/
You can also see the thing I'm talking about in steveo1kinevo's video:
https://youtu.be/GZv-5RBdugM?t=823
Look at how he changes power settings for the cruise. He flies at following power settings: around 86 % / 2000 RPM. Then he reduces RPM to 1900 and gets around 90 %. Do the math - you will get 172000 from both 86*2000 and 90*1900. Actually, you will get 172K and 171K but this difference is far below the accuracy of me reading his gauges.

PS
Let me say that apart of this issue your DHC-3 is a joy to fly and I really love the work you put into correct simulation of this aircraft and it's engine!