Am sure this has been asked before, but cant find it with searches.
The model is a pylon mounted pusher prop design and I am trying to come up with a good starting point for the thrust angle of prop.
Is there any way that you can resolve vectors from motor angle to CG, or CoM or whatever to determine an approximate thrust angle?
The model in question has a Clark-Y wing with 250mm chord and the prop is about 300mm behind the CG and raised 100mm above the chord line.
Cheers,
Glenn.

The only thing that can counter the pitching moment from a pylon motor installation is the horizontal stabilizer. The thrust line should point at stabilizer so that whatever your power setting the stab will produce an equal and opposite reaction. like this:

--Norm

Thanks Norm,
Its not too late to raise the height of the horizontal stab, so might give that a try. Intuitively, the raised height of the prop must induce a pitch down with throttle, so might raise the prop end of the motor shaft for a bit of compromise, as other pushers seem to do this.
Cheers,
Glenn.

I was under the impression the nose of the pusher motor needed to angle slightly down. The counters pitch up under full power.

Frank

Some more threads concerning this. I haven't had the chance to read them yet. This questions has always made me wonder also since I have built and flown a number of pushers.

http://www.rcgroups.com/forums/showthread.php?t=76309

http://www.rcgroups.com/forums/showthread.php?t=562798

http://www.rcgroups.com/forums/showthread.php?t=642029

http://www.rcgroups.com/forums/showthread.php?t=888546

http://www.rcgroups.com/forums/showthread.php?t=908829

http://www.rcgroups.com/forums/showthread.php?t=656724

Frank

Some interesting threads there, I only searched this one. No real definitive solutions though!
Cheers,
Glenn.

Originally Posted by murocflyer

I was under the impression the nose of the pusher motor needed to angle slightly down. The counters pitch up under full power.

Frank

Normally that's what you'd do but a pylon mount raises the thrust line so high that pointing it at the CG would require a very large angle, turning too much power into downthrust. Pylon mounting leads to bad compromises.

What about a 4 or 5 degree nose down attitude?

Frank

It would be best to have the thrust line parallel to the free stream but there's actually very little thrust lost because of a few degrees of offset. This was explained to me years ago by a guy building rocket planes. P-factor and gyroscopic precession make prop behavior more complex but basically they only contribute a 90 degree offset in the plane of rotation and vibration. ignore them for the thrust analysis. Draw a vector diagram with the thrust line as the hypotenuse. As you can see from this example the propulsive force is hardly diminished at all at small angles. But, what's really weird, is that a relatively large force perpendicular to the propulsion appears. Pleas don't ask me to explain the math behind is because I can't. It looks like you're getting something from nothing but you're not. It has something to do with the difference between energy, force and power. Although the two forces add up to more than 100% the total energy hasn't changed.

--Norm

Originally Posted by nmasters

But, what's really weird, is that a relatively large force perpendicular to the propulsion appears. Pleas don't ask me to explain the math behind is because I can't. It looks like you're getting something from nothing but you're not. It has something to do with the difference between energy, force and power. Although the two forces add up to more than 100% the total energy hasn't changed.

--Norm

No complex math involved; it's simple trigonometry. (98.5^2+17.4^2)^(1/2) = 100.

Dan

Yeah, the hypotenuse of a right triangle is always shorter than the sum of the other two sides. I just wouldn't have believed that forces could be worked out this simply until a couple of rocket scientists told me so.

Originally Posted by nmasters

But, what's really weird, is that a relatively large force perpendicular to the propulsion appears. Pleas don't ask me to explain the math behind is because I can't. It looks like you're getting something from nothing but you're not.

Precisely. Because the force perpendicular to the direction of motion does not perform any work, thus no energy is converted. Just like blind power and effective power in an AC cirquit.

True, as long as the plane is moving horizontally, the vertical component of thrust isn't doing any work. But with the rocket illustrated it is adding that 17.4% of the force to the weight of the airplane.

don't confuse the decomposition of forces as actual forces. Vector sums are just a mathematical tool to operate calculations on complex force interactions. Some of the assumptions made do not translate directly in reality, like having all vectors attached to the same point on a rigid object. Essentially, you can put two identical motors on a rigid object, set up to pull in opposite directions. The resultant vector will turn out to have 0 length, but there will still be a force there trying to pull the object apart. As a whole it won't move, but whether it explodes is another matter
And approaching this in the opposite way, an object with no forces applied to it could be correctly represented in a vector system as
<---infinite length---(object)---infinite length--->
and yet it won't have any force applied to it, let alone forces with an infinite strength

Originally Posted by Brandano

don't confuse the decomposition of forces as actual forces. Vector sums are just a mathematical tool to operate calculations on complex force interactions. Some of the assumptions made do not translate directly in reality, like having all vectors attached to the same point on a rigid object. Essentially, you can put two identical motors on a rigid object, set up to pull in opposite directions. The resultant vector will turn out to have 0 length, but there will still be a force there trying to pull the object apart. As a whole it won't move, but whether it explodes is another matter
And approaching this in the opposite way, an object with no forces applied to it could be correctly represented in a vector system as
<---infinite length---(object)---infinite length--->
and yet it won't have any force applied to it, let alone forces with an infinite strength

I'm not quite sure what you're talking about. The vector sums are not merely a mathematical tool. In the case we're discussing, there actually is a 17.4N force downward and a 98.5N horizontal force. Your discussion of internal forces is a different matter entirely, but even when you're dealing with those force decomposition is completely valid.