Can you explain why the prop must be tilted up?

From Don Stackhouse:

Uh, oh, now you've done it! You've invited me to submit another answer to an engineering question! And of course, why would I want to submit a short answer that only applies to this particular case, when with a little more work I could give an answer that explains the entire subject? ;-)

Martin Irvine writes:

...I had time to reconsider my last posting. It is true only if the prop is ahead of the centre of lift. If the pusher is behind this,as it usually is, then the motor must be tilted "up", to push the nose up, (I think). At this point I am rather less certain of what I wrote. Maybe Don S. or Scott can contribute?

Martin, your original post was pretty much correct, but it appears that some of your reasons might be a little confused. Center of lift has very little to do with it, but I'm getting ahead of myself a bit.

Oh, boy, now we need to talk about something dear to the heart of most engineers, the subject of "vectors"! Many of you are probably already familiar with these. Some of you may have even developed an intense disliking for them, perhaps from some previous unpleasant academic misadventures, but for those of you who haven't yet been introduced, I'll start with a brief (yeah, right!) primer. Some of this explanation is needed to understand thrust line theory, some will probably be required in the future for other topics; I might as well give you the whole thing. Don't worry, I promise to (eventually) get back to the original subject of thrust lines!

In the engineering world there are, generally speaking, two kinds of quantities, "scalars" and "vectors". Scalars are simply a quantity, such as mass or area. Vectors describe something that has both a quantity (or "magnitude") and a specific direction, such as weights, velocities, and forces. Thrust is a vector, as is lift and drag.

In engineering analysis we usually represent a vector quantity graphically as an arrow. The direction of the arrow indicates the direction of the vector quantity, and the length indicates its magnitude.

If you have several vectors of the same particular type (such as a collection of forces, or a combination of velocities), all acting on the same object (our model, in this case), you can graphically add them together to find the magnitude and direction of a single vector that represents the combination of all of them together. Re-draw them in a chain, with the tail of the second arrow starting at the nose of the first, and the tail of the third starting at the nose of the second, and so on till the chain is complete. Now draw an arrow from the starting point of the chain (the tail of the first arrow) to the end of the chain (the nose of the last arrow). This new arrow is the vector that results from the combination of all of the arrows in the chain, so therefore we call it the "resultant".

An example of this is the description of an aircraft in a coordinated turn. The aircraft is at some given bank angle, with weight pulling down and with centrifugal force (ok, for all you purists out there, I know I'm supposed to call it "centripetal" force, but please indulge my bad habits anyway) that comes from the aircraft's turning flight path pulling out horizontally to the side. Both of these two forces act through the C/G. If we re-draw the centrifugal force vector with its tail starting at the nose of the weight vector, then draw a new vector from the C/G (the start of the weight vector) to the nose of the centrigugal force vector, we get a vector that represents the total load on the wings. Please notice two things about this "resultant": It is longer than the weight vector (in fact the ratio between the length of the resultant and the length of the weight vector is the "G" load in the wings), and it is PERPENDICULAR to the wings if the turn is "coordinated" (i.e.: the bank angle is correct for that particular turn). Since the airplane cannot distinguish between cetrifugally created forces and gravity derived forces, and the resultant is perpendicular to the wings, the aircraft "thinks" it's in level flight, but with a higher weight. This is why, after you roll to the correct bank angle for a turn using ailerons or rudder, you center the stick (because the now turning model thinks it's in level flight) and pull in a small amount of up elevator (because the model thinks it's heavier, and therefore needs to make more lift).

The reverse is also true; you can also break a vector down into "components", any combination of vectors that when added together nose-to-tail result in the original vector. For example, we could calculate the lift vector for a wing, then break it down into its components in the vertical (weight) and horizontal (cent. force) directions.

Now at this moment, I'd like to introduce the subject of "moments", or twisting forces. There are two basic kinds. A "torque" is a force vector acting perpendicular to a lever arm (or "moment arm"). It creates a twisting force in addition to the vector's linear force. A "couple" is a pure twisting force, without any associated linear force.

We see examples of both of these in a typical helicopter. The power from the engine, applied to the main rotor shaft and absorbed aerodynamically by the main rotor blades, creates a "couple", or pure twisting force, that tries to turn the helicopter away from the desired heading. We counteract this with a "torque", the thrust of the tail rotor (the force vector) acting at the end of the tail boom (the moment arm). This cancels out the couple in the main rotor shaft, so the heading now stays the same, but it leaves us with a sideways force (the thrust of the tail rotor) which is trying to push the whole aircraft sideways.

To counteract this, we bank the helicopter slightly the other way. This tilts the lift vector from the main rotor to one side, so that the main rotor's lift vector now has a horizontal component equal and opposite to the thrust of the tail rotor. This stops the sideways drifting effect. Of course since we're now using some of the main rotor's lift to make sideways force, and the helicopter still weighs the same, we have to slightly increase the lift of the main rotor, which increases the required twisting force in the shaft, which increases the thrust required from the tail rotor, which increases the required bank angle, which... By now you're probably beginning to see just why helicopters are so complicated to fly!

If we extend the arrow representing a vector into an infinitely long line, we have the "line of action" of that vector. In most cases, as long as the magnitude of the vector stays the same, the location of that vector can be anywhere along that line of action without changing the way that vector influences the system it's acting upon. When calculating the moments created by any particular vector, the moment arm is the perpendicular distance from the pivot point of the system (for an aircraft in flight, this is normally the C/G) to the vector's line of action.

(End of part 1, continued in following post)

Don Stackhouse @ DJ Aerotech
djarotec@bright.net
http://www.bright.net/~djwerks/