Chuck Beaty...could you verify my tail rotor thrust calculations

[StanFoster]

In my Translating Tendency thread....I was displaying my figures for what my Helicycles tail rotor would have to exert to counter the torque driving it and the main rotor. I would like someone like Chuck Beaty, or anyone that can display the math here.... to come in and show how much my thrust in pounds my tail rotor needs to counter the torque in the following scenario...............

My turbine is derated to 100 horsepower. The scenario is that all 100 horsepower is being consumed by my main rotor and the tail rotor. I am not saying it takes 100 horses to hover...but if I was max loaded...and whatever it took to consume 100 horsepower .

The main rotor turns 620 rpm's and the tail rotor is taken off the same gearbox. If I am hovering max load...and whatever it takes to put 100 horses into the gearbox that drives the main and tail rotor....how many pounds of thrust does the tail rotor need to counter this 100 horses?

The tail rotor's driveshaft is 12 feet from the rotorblade driveshaft.

I came up with 847 foot pounds of torque going into the main rotor/tail rotor drive. I arrived at this by taking one horsepower , which is 550 foot pounds/sec....times 100. That is 55,000 foot pounds/sec. I then figured the circumference of a one foot radius circle....which is convenient for figuring foot pounds over distance. Then I figured the rotor rpm at 620, and this means that in one second...the rotor turns 10.33 turns. With a circumference of 6.283 feet....this means the distance the force at one foot radius moves is 64.92 feet. I then divided 55000 by 64.92 and arrived at 847.19 foot pounds of torque the rotor and tail rotor are taking as they consume the 100 horses.

My thinking that since the center of the tail rotor is 12 feet out....the force it exerts to counter this 847 pounds of torque would be one twelth that, or 70.59 pounds.

Now....

My figures and my neck are on the line.....so please correct my thinking if I am figuring this wrong. I want to learn....and wont mind admitting I made a big mistake learning.....but I just want to know if I am right....or why I am wrong. Thanks.


Stan

[kolibri282]

Stan,

to calculate the torque your rotor absorbes you have to integrate the drag force along your rotor blade since not all of the drag forces act at the tip of your main rotor. Could you please list your main and tail rotor diameter and rpm as well as gross weight, one could then calculate an approximation. (but that's for the weekend since it's two minutes to midnight over here and I'm fading fast..;-)

[raghu]

if you assume all of the 100 hp is absorbed by the rotor, then the torque generated is given by power*5252/RPM = 847 ft lb. Assuming a 12 ft length from the CG for a tail rotor, the thrust required by the tail rotor to balance is 847/12 = 70.5 lbs

[C. Beaty]

Say the tail rotor takes 10 hp and the main rotor takes 90 hp.

Torque = (hp x 5252)/rpm
(90 x 5252)/620 = 762 ft-lb (main rotor torque)

Tail rotor thrust = main rotor torque/moment arm
762/12 = 63.5 lb.
******
100 hp at 620 rpm would indeed be 847 ft-lb of torque. But the twisting effort of the main rotor depends on how much power it consumes; tail rotor power doesn’t count against the main rotor.

The factor 5252 comes from 550 lbs, one ft/second, 60 seconds per minute and 2pi. You can work out the sequence.

[raghu]

Stan basically your problem, as stated, has two variables and one equation; the only way to solve it is to assume one variable. I assumed (as indicated in my post) all the power going to the (main) rotor which gives you an upper bound of the tail rotor thrust, while CBs assumes a 90-10 split. You can assume other variables and solve as well- for instance the RPM of the tail rotor.

[raghu]

Originally Posted by StanFoster

Chuck- please help me refine my thinking. Ok, you are saying if the main rotor consume 90 horsepower, and the tail rotor consumes 10 horsepower, why doesnt the tail rotor also have to supply thrust to counter the torque being added to the main rotor. I mentioned that the main and tail rotor were consuming 100 horsepower. Thanks for your input. Stan

Stan, the main rotor and tail rotor torques twist the helicopter in different planes-yawing and pitching plane respectively- hence they are not additive. The Main rotor torque (induced in the yawing plane) is countered by the tail rotor acting along the tail arm. The tail rotor, of course, also induces a twisting force but it is in the pitching plane (Pitches the craft up or down) which is countered my the main rotor thrust being slightly ahead (or behind) the CG.

[C. Beaty]

Only the torque applied to the main rotor is trying to rotate the airframe in the opposite direction.

If the main transmission had provisions for driving generators and hydraulic pumps, the torque applied to them wouldn’t try to rotate the airframe and would not require compensation. Accessories like that are closed systems and don’t react with the outside world.

If your transmission had a power takeoff shaft driving a sawmill, then you would have a torque reaction to contend with if the sawmill was not bolted to the airframe.

[gyronutjoe]

What about drag of the main rotor going through the air?

[Jean Claude]

Stan, your calculation is correct. Chuck added a small correction by subtracting the tail rotor power (10%). I adds also a small correction by subtracting the power losses in the main gearbox and V-belts (about 5% ?): If the power of the turbine is 100 hp, then only 100 hp -10% - 5% reach to the main rotor.
Jean Claude

[raghu]

Originally Posted by C. Beaty

Only the torque applied to the main rotor is trying to rotate the airframe in the .

CB, I am not sure I understand. Are you saying only the main rotor torque induces a reaction on the airframe? the tail rotor torque induces no external reaction on the airframe- akin to pushing a car while sitting in it?

[C. Beaty]

Of course the tail rotor produces a reaction on the airframe, Raghu, but as you’ve previously stated, about the pitch axis. Air is the outside world.

[raghu]

Thanks CB! Was not doubting the 'ol' master of mechanics'..... just perplexed by what you where trying to say in post 11 .......

[C. Beaty]

Raghu, I was trying to plagiarize an example from a thermodynamics class I took ~100 years ago –well, not quite but it seems that way- but have forgotten exactly how it went. Had to do with a closed room (not Maxwell’s demon).

Things like generators and hydraulic pumps don’t react with the outside world and can produce no unbalanced torques any more than you can lift yourself by your own boot straps.

Things that do react with the outside world product torque reactions; wheels, oars, propellers, etc.

[Arnie Madsen]

Originally Posted by gyronutjoe

What about drag of the main rotor going through the air?

I was thinking the same thing Joe. In crisp cold air a high blade pitch at high engine power would produce the most blade drag and the most torque . The tail would also have the advantage of the high air density ..... so does everything balance out ..... ??

..... compared to a 100* Florida day with thin air , less torque , and weaker tail rotor response .... ???

Days like this I am glad my math skills are weak and I do not have to calculate it. thanks everybody for some answers.

[kolibri282]

The tail would also have the advantage of the high air density ..... so does everything balance out ..... ??
..... compared to a 100* Florida day with thin air , less torque , and weaker tail rotor response .... ???

Here are two complete trim solutions for a UH-1H at 60 knots, one for standard atmosphere and the other one for an air density which is 10% higher (this value is arbitrary just to show the effect)
The main and tail rotor collective (thtaN/thtaTR) angles are slightly smaller for the elevated air density, as one would expect, but due to the higher drag (air density enters linearly in aircraft drag) the main rotor thrust (fMRres) is slightly higher at 27461N (comapred to 27455N ) and therefore main rotor torque (QMR) is also slightly higher. Note that all force and moment components are in body coordinates. The pitch trim angle (thta) is slightly smaller for the elevated density due to the higher H-stab pitch moment


rho = rho standard
thtaN_____13.52°____thtaTR_____2.63°___fMRres_2745 5.95___QMR__5084.04
fMR(1)__________1282.67___fMR(2)__________-598.06___fMR(3)________-27419.45
fTR(1)_____________0.00___fTR(2)___________589.49_ __fTR(3)_____________0.00
fFus(1)_________-950.14___fFus(2)___________-0.00___fFus(3)___________-4.48
fFN(1)___________-43.61___fFN(2)____________-0.01___fFN(3)_____________0.00
fHS(1)____________-8.25___fHS(2)_____________0.00___fHS(3)___________ _23.40
fCG(1)__________-280.67___fCG(2)_____________8.58___fCG(3)_________ 27400.53
fR(1)___1.3926638e-011___fR(2)___5.5049298e-012___fR(3)___4.7293724e-011

Moment Equilibrium
mMR(1)_________-1237.97___mMR(2)__________-217.25___mMR(3)__________5030.87
mTR(1)__________1237.97___mTR(2)_____________0.00_ __mTR(3)_________-5030.93
mFus(1)____________0.00___mFus(2)___________76.75_ __mFus(3)___________-0.00
mFN(1)_____________0.00___mFN(2)_____________0.00_ __mFN(3)_____________0.06
mHS(1)_____________0.00___mHS(2)___________140.50_ __mHS(3)_____________0.00
mR(1) 1.0686563e-011 mR(2) -1.0174972e-011 mR(3) -4.0927262e-011

Trim_phi_LFrA_______0.02°____thta_______0.59°____p si_______0.00°

rho_=_1.1_rho_standard
thtaN_____13.11°____thtaTR_____2.58°___fMRres_2746 1.32___QMR__5405.80
fMR(1)__________1276.20___fMR(2)__________-635.90___fMR(3)________-27424.28
fTR(1)_____________0.00___fTR(2)___________626.79_ __fTR(3)_____________0.00
fFus(1)________-1045.23___fFus(2)___________-0.00___fFus(3)___________-2.29
fFN(1)___________-47.97___fFN(2)____________-0.01___fFN(3)_____________0.00
fHS(1)____________-8.98___fHS(2)_____________0.00___fHS(3)___________ _25.16
fCG(1)__________-174.02___fCG(2)_____________9.12___fCG(3)_________ 27401.41
fR(1)___1.1056045e-011___fR(2)___4.5616844e-012___fR(3)___7.2759576e-012

Moment_Equilibrium
mMR(1)_________-1316.31___mMR(2)__________-203.43___mMR(3)__________5349.26
mTR(1)__________1316.31___mTR(2)_____________0.00_ __mTR(3)_________-5349.30
mFus(1)____________0.00___mFus(2)___________52.35_ __mFus(3)___________-0.00
mFN(1)_____________0.00___mFN(2)_____________0.00_ __mFN(3)_____________0.04
mHS(1)_____________0.00___mHS(2)___________151.08_ __mHS(3)_____________0.00
mR(1)___8.8675733e-012___mR(2)__-2.5579538e-012___mR(3)__-3.0922820e-011
Trim_phi_LFrA_______0.02°____thta_______0.36°____p si_______0.00°

PS: the underscores are necessary since the software deletes all blanks and tabs so you can't format a table with blanks in this forum