How Does Upwards Drag Force Create Horizontal Torque in a Paper Helicopter?

[asu04]

I'm working on a Paper Helicopter Experiment in school where I'm looking at the deflection angle of the rotors versus the rotational velocity of the copter. I'm having difficulty forming a relationship between the deflection angle and the torque that the helicopter experiences

But how exactly does the upwards drag force resolve into a horizontal torque? I've been drawing some free body diagrams and the only way that I've seen that torque could have been produced was if the blades pushed back on the air with a force equal to the component of drag force parallel to the slant angle of the wings. Is this the reason or is it something else?

Any help much appreciated

By the way if anybody is wondering what a paper helicopter is:
http://www.theonlinepaperairplanemuseum.com/AZMuseum/R/RotoCopter/Robo-CopterPic.JPG

Looks like a sycamore seed.

Thanks

 

[hikaru1221]

I think the drag force is just the net (or total) force. Each blade experiences a force which has both horizontal and vertical components due to the air. Because the vertical components of the 2 forces on the 2 blades cancel each other, the net force is the sum of horizontal components.

I haven't figured out the exact equation for the forces, as the air's motion is complex.

 

[asu04]

Hmmm...on a related note if I knew the angular acceleration of the helicopter how can I compute its moment of inertia to obtain the net torque on the copter? I heard that for shapes with simple objects, the moment of inertia can be found by considering its dimensions?

 

[hikaru1221]

The moment of inertia (I) always has this form: I = k x Mass x Square of dimensions, where k is some coefficient depending on the shape and Square of dimensions is a sum depending on the dimensions and the shape. So we always have to find the dimensions to calculate I, however simple or complex the shape is.

For the paper helicopter, you can find I about the central axis of each blade and the body by assuming they're rectangular and their mass is uniformly distributed. It's easy for the body (it's totally similar to I of a rod about its axis of symmetry). For the blades, first, you should find I about the axis going through its center and parallel to the central axis of the helicopter, and then, apply Huyghens-Steiner theorem (see http://en.wikipedia.org/wiki/Parallel_axis_theorem).

 

[rdl]

for sharing your experiment with us! It's great to see young scientists exploring the principles of flight through hands-on experiments like this.

To address your question about the relationship between deflection angle and torque, it's important to first understand the forces at play in a paper helicopter. As you mentioned, the upwards drag force is created by the rotation of the rotors, and this force must be balanced by the weight of the helicopter in order for it to stay in the air.

When the rotors are spinning, they create an angular momentum that causes the helicopter to rotate in the opposite direction. This rotation creates a torque, or a turning force, which can be measured by the deflection of the rotors. The greater the deflection angle, the greater the torque being exerted.

To answer your question about how the upwards drag force resolves into a horizontal torque, you are correct in thinking that it is due to the blades pushing back on the air with a force parallel to the slant angle of the wings. This force, called the lift force, is what allows the helicopter to stay in the air and is also responsible for the torque that causes it to rotate.

I would suggest continuing to experiment with different rotor designs and rotational velocities to see how they affect the deflection angle and torque. This will help you better understand the relationship between the two and potentially lead to further discoveries.

Keep up the good work and keep exploring the fascinating world of flight!