I'm building a boomerang for a project, but I've made it a goal for myself to have a boomerang which can fly at lest 200 feet, so how would I increase the flight path. Angular momentum increases in the direction of the torque, but why? Is there a way I can increase the angular momentum through a longer path? please help, thank you for your time.

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    $\begingroup$ It would be helpful if you showed the equations for boomerangs that you have found so far. 200 feet is a considerable distance - to overcome the air drag for the round trip you clearly need a lot of kinetic energy, so presumably a heavy boomerang and a strong arm - and sufficient angular momentum to maintain the lift during the path. "Big and heavy" is a first step. $\endgroup$ – Floris Dec 29 '14 at 4:24
  • $\begingroup$ But if I increase the area wouldn`t I have to increase the force I apply to increase the lift, plus more weight means more drag. $\endgroup$ – Ben Dec 29 '14 at 4:37
  • $\begingroup$ Drag is a hydrodynamic property that is completely independent of the weight (what you really mean is mass) of an object. It only depends on the shape. I think what you want is higher moments of inertia. Moments of inertia can be increased (for a given mass) by placing the mass further outside while leaving the center of gravity the same. I would suggest a tungsten tipped carbon fiber design. Good luck! $\endgroup$ – CuriousOne Dec 29 '14 at 5:33

A full exposition of the equations of motion of a boomerang can be found at this link - it is really quite hard work to read through. Let me try to summarize the main points:

  • A boomerang derives "lift" from the winged shape and its rotation
  • The boomerang returns because the lift is asymmetrical: when you add the velocity of linear motion to the velocity due to rotation, you can see that the leg that is moving forward is experiencing greater lift than the leg moving backwards
  • The differential lift results in torque on the boomerang, and will cause it to start tilting; this results in an approximately circular path
  • If you throw the boomerang so it has a vertical component of velocity, it will lose linear speed as it "climbs" - this makes the flight path less circular

So what do we learn for getting the boomerang to go far? First - we need to slow down the rate at which it turns in its path: this means that the linear velocity needs to be small compared to the wing velocity (less differential lift), so we need fast rotation and a wing with small angle of attack. Second, we need a large moment of inertia so the boomerang maintains rotation during flight - @CuriousOne's suggestion of a tungsten tipped blade is certainly interesting (consider a 3 legged boomerang to give a more favorable construction). Finally - throw it hard.

But I think that keeping the lift small and the rotation fast are the keys. Do google "equations of motion of boomerang" - you will find quite a treasure of papers with more detailed analysis. Probably the most accessible are http://plus.maths.org/content/unspinning-boomerang and http://large.stanford.edu/courses/2007/ph210/moon2/

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    $\begingroup$ but how are you a reliable source I can cite in my bibliography? $\endgroup$ – Ben Dec 30 '14 at 6:58

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