a simple propeller on a common engine , turns one way , the engine tries to react the opposite direction.
Does the propellor produce a greater twisting force than a propulsive force?
Is there a fixed ratio?
which direction of force is the greatest?
Too easy?

hmmm,
How about, "For every action there is an opposite yet equal reaction?"

Just thinking ....
If you put a pusher prop on the rear of the engine would this qualify as a "counter rotating prop set up"?

The fuel and throttle set up would be interesting.

Who wants to pull the pull the glow starter off of this set up?

But will it take off?

lol

Yep, it's all about Newton's third law of motion. The prop spinning clockwise tries to roll the airplane counterclockwise along the longitudinal axis. It's called the reactive force.

Glad to see you're giving your brain a workout. Some people have no curiosity and it drives me nuts. (short drive)

This hobby is supposed to be fun right?
Sometimes we all get to serious

OK -guys -- the queston was tho -- is the twisting force, greater that the propulsive force.
sorry -it was a trick question.

And you sneaky devil.... It's a trick because the propulsive force is a MUCH different force than the twisting force.

And no question is really a trick question. But some will trick the people who don't read the words for what they say.

ACTUALLY...............

I would suggest this answer to your trick setup. (If there can be no trick questions, then must there be no trick setups? hehehehe)

I would suggest this........

If the prop you placed on the rear truly would counteract the torque of the engine/propeller on the front of the engine THEN
Neither prop would turn. And it would be quite easy to remove the glow starter.

hmmmmmmmmmmm or would the engine be turning and neither prop moving!!!!!!!!!

OH MY GOD............. it IS a trick!!!!!!!!!!!!

I was bored
Way back when-- I was a draftee and ended up at a NIKE base -taking care of the stand by generators, compressor, lawnmower etc..
We had a TDY 2nt Lt who was shaping things up for an upcoming IG inspection.
Nice guy.
He suggested we paint the floor in the Generator shack- to make things "stand tall".
I tactfully reminded him that as our paints were conductive, it could possibly cause a problem if the floor flooded .
He said "of course , better just clean em up".

I was the guy who would have done the painting --
What the he-ll --I got away with it ---.

If your plane flies and it flies straight, all the twisting force (T) doesn't matter because your propulsive force (rpm on the prop) is definitely greater. If your prop has the same mass as the plane, then ****etaboutit.

Question is, what happened to the feet?

If Torque is measured in ft/lbs and thrust is measured in lbs, what happened to the ft?

What do you mean about propulsive force? What do you mean about twisting force? To me, twisting force means torque measured by (HP(5252))/rpm). Propulsive force could mean the same thing since the engine is doing work. If you meant propulsive force made by the propeller, then it would be a different equation T=CsubT[(air density*rpm)^2*D^4). CsubT is the thrust coefficient of the prop.

Anyway....musings.

nibble nibble swallow.............

You hooked one.

Hey, are you saying that you are proud of having put one over on a 2nd LT?

BTW, if you had simply grounded all the electrical devices to the roof, then flooding would not have been a problem, right? Didn't that 2ndLooie know nothin?

For static conditions it would depend on the prop pitch I would think. In flight the ratio would be all over the place I'd think. Or perhaps it wouldn't.

You had too much coffee just before you typed that didn't you.....

So easy, even a cave man can figure it out.


HUH?

two for two----
not bad fishin----

From my observation spot on the stream bank, I'd say you've definitely hooked 2. Two of 4. Two were just us kibitzers.

Neither. The engine produces the twisting force, and the propeller produces the propulsive force. The twisting force will be the greater of the two, because the propeller is converting the twisting force the engine is producing and turning it into thrust and there will be mechanical losses in this conversion. Just a guess.

and the winner is--------hdsoar!!!!!!!!!!!!The question was a simple logic question.
I expected some formulae-- etc., but you nailed it.

As you obviously know, any conversion from device to device or mechanical ratio or change of direction etc., has a resultant loss of some kind -NEVER a gain .
The perpetual motion guys have tried n tried but it still ends up - a loss for every change.

I'm not sure it works that way here. Let's look at your favourite flat foamies which I've also started flying in the quest to learn to prop hang/hover.

The model weighs around 8 to 10 oz as I recall. But to counter act the torque while hovering at my 8 to 10 oz of thrust I only need to use perhaps 15 degrees of aileron. I'd be very willing to bet that the prop blast over that small area at a 15 degree angle isn't anywhere near an 8 to 10 foot/oz of torque.

Just for the heck of it I plugged some numbers that may or may not make sense into a servo torque calculator. I used 30 mph as an outside number for the prop blast speed, 15 degrees surface deflection on a 2.5 inch wide by 6 inch long surface (the part in the prop blast) and 15 degree max servo deflection so it would indicate a 1:1 control horn to servo arm overall. The servo load calculation came back as only 1.22 inch oz of torque. Double that for two surfaces and it's still only 2.44 in-oz of torque. Somehow that doesn't seem to correlate with a prop that's generating 8 to 10 oz of thrust.

So.... I went to Foilsim, my favourite airfoil lift program for playing, and plugged some numbers in....

A 2/3:1/3 wing:aileron at 15 degrees with a chord of 9 inchs forms a roughly equivalent flat arc section of 5% and 5 degree angle of attack. With my mythical 30 mph prop blast blowing over it this panel will generate JUST OVER ONE LB OF LIFT ! ! ! ! I would not have thought it would be that high. Now of course all this is based on the blast being at 30 mph. I found an Emotor calculator that makes it look like 30 mph is a little high. At what appears to be a more realistic 25 mph prop blast at the throttle setting to hover the wing segment in the blast appears to be generating just under .8 lbs of lift to hold the model from spinning or a total of 1.57 lbs of lift to prevent a hovering model of around 10 oz from spinning.

So it would appear that we do indeed have a winner and that it takes more "lift" to hold the model from torque rolling than it does to keep it hovering.

Now mind you if I'm way off on my camber and induced angle of attack then it may not all fit. For example if we reduce the deflection to 7 degrees that reduces the camber and AoA to 3% and 2.5 degrees. At the same 25 mph prop blast we now have the two sides adding up to a total of just over .8 lbs of lift for a more or less 1 to 1. A big change.

But barring some real numbers that we can use to get better answers it would appear that the lift needed to counter the torque related to the thrust is more or less equal to or exceeds the force of the thrust.

The same question bugged me in flying /hovering etc..
The force which would instantly accelerate the model -seemed higher than the force which tried to rotate the model.
The thing I kept coming back to ?
th accelerative force ^is derived from the twisting force >.
Y/N?
This would lead me to believe that any calculation which showed the derived force as greater - has to be flawed
OR--- it really is possible to get "more out of it than you put into it".
So far - I have not seen that happen.

Only problem with that interpretation is you're watching the rotation of the model as a measure of the force.

The aerodynamic "power" of the model has a great deal to do with it's rotation.

A big foamy with lots of areas will rotate less than a little foamy. Neither are flying with any "power" in the aero parts, but those areas still resist being pushed around in all that air.

Put a 25kW motor on a big foamy. Hover it and then try to accelerate. If the foamy is big enough, you don't see much torque roll right? Might not even see one at all. Put a 150kW with it's appropriate propeller on the same foamy. Or trim off some of the foamy wing span and fuselage and keep the 25kW. And what do you see?

I'd say this test method doesn't suit the hypothesis. and change my observation to suggest that the angler hooked his shirt on the first cast

Yeah, I misunderstood the question somewhat. I guess I assumed that he was digging up that age old question of torque.

The sick part is, that is fun for me!! [sm=spinnyeyes.gif][sm=spinnyeyes.gif]

A 25 kilowatt foamie?
A 150 kilowatt foamie ?
Dear me!!!

anyway you just got snagged on the cast -- The original "question" was
which is greatest force. as one force is a product of the other - the second can never be more than --or even equal to .
(Is one foot pound of torque equal to one pound of lift?)

Dick,

The question doesn’t make sense. Twisting force isn’t a force at all, but a torque. A torque is the product of a force acting at a moment arm. For a given torque, the force can be any value depending on the moment arm, whereas the propulsive force is simply a force.

Your statement that, “…any conversion from device to device or mechanical ratio or change of direction etc., has a resultant loss of some kind -NEVER a gain .” is an attempt to state the law of conservation of energy but is not completely accurate. Using a simple lever it is quite possible to gain a large force output from a small force input.

Applying energy conservation to the situation described in your question:

work input to the propeller (by the engine) = work done by the propeller (in moving the air) – loss

The work input would be the torque applied to the propeller times the RPM.

The work done by the propeller would be the mass of air affected by the propeller, times the change in velocity across the area of the propeller per unit of time.

While it is obvious that work in must equal work out minus loss, for airplanes of conventional proportions, the force required to produce the torque to turn the propeller is much less than the thrust force produced by the propeller.

yes --true -- only one thing to add: the lever comparison - is a force & distance relationship ----(trick question again)
I said it was a trick question--- your last paragraph (sentence) says it as it is.
I deliberately muddied it up.
In setting up engines to produce best results for racing or other applications -- it is very common for people to try to divorce torque from horspower.
Often, they are described as two seperate animals -one for pulling hard the other for going fast. The idea that torque is required to get horsepower is manytimes , missed.

Wot?



old git - - - - - aka John L.

starting with a spinning shaft --the torque required to spin a load at a speed is expressed as horsepower
torque is not a function of time -
rpm is