i wanted to ask if the size of the wheel is important? currently i have a setup of 4 motors geared to high speed going to a 7:1 ratio, then the wheels are the 2.75" small rubber wheels. So far i am getting no distance what so ever. I just wanted some explanation and some science behind what is happening with the wheels and what i can do to improve it.
Thanks for replying
The short answer is yes. There is a lot to consider. You need to gear up the motors quite a bit.
Check out:
https://vexforum.com/t/physics-of-the-flywheel-launcher/29357/1
Also:
The mass of the wheel will be more important than the radius. The mass with the angular velocity of the wheel drives a bunch.
the size of the wheel is definitely important, if you want I can post some spread sheets I made, I have ones that will tell you what launch velocity you need for any given launch conditions and another that allows you to mess with launch conditions and it wills show you the ballistic trajectory.
to answer your question: yes, wheel diameter does matter, the larger the diameter the faster the outer edge of the wheel goes, the faster it will accelerate the ball and the faster it will go. the calculations I have mandate at 26ish FPS launch velocity at 60 degrees, for full court shooting. this works out to a 9.8 to one ratio if using 4 inch wheels and speed motors.(I can give you that spread sheet too) all that math is theoretical and does not include friction, but the 9.8 to one allows for about 25% error in friction and will still make the goal
Obviously the size of wheels is important. Consider how using different wheel sizes affects the speed of your robot’s drive. Using larger wheels will cause your robot to cover a greater distance for every rotation of the wheels, meaning the robot will travel at a higher speed. For the same reason, using larger wheels on a launcher will cause the ball to be launched farther.
Additionally, there are other reasons why wheel size is important, involving the amount of contact between the ball and the wheels. My suggestion would be to experiment as much as you can, and see and take note of how every change you make to your launcher affects the trajectory of the ball.
Sorry I should have been more verbose. The radius of the wheel does give it a ton more speed as radius is increased but a larger mass will keep not slow the wheel down as much between shots.
Try making the same size wheel heavier and see what happens.
Kinetic energy is 0.5mr*w so radius and mass have the same factors.
I would love to see your spreadsheet. Please post it.
Yes the radius matters, for two main reasons.
The first is energy efficiency. Larger wheel radius means the rim of the wheel will be going faster, so the ball will go faster. But that doesn’t matter, because you can also achieve a higher speed using gear ratios. But it actually does matter, because a larger radius means less gearing which is more efficient. With large enough wheels you can probably achieve the speed you need with turbo gears and a single reduction, which is quite a bit more efficient than using a double reduction.
The second is mass efficiency. Within sensible limits, it’s good for your flywheels to have a lot of angular inertia. This means they can store rotational energy, which is transferred to the balls as they are thrown (if your rollers aren’t storing rotational energy then they aren’t flywheels - they’re just wheels). Angular inertia increases with mass, and also with the distance between that mass and the centre of rotation. That means a wheel with a larger radius can store more rotational energy with less mass.
The tradeoff, of course, is that larger wheels take up more space.
It depends on the moment of inertia of the rotating wheel (which is not simply proportional to the radius). Also the relationship to rotational velocity is quadratic rather than linear.
Overall the equation for kinetic energy of a rotating wheel is somewhere between 0.25mr^2*(omega)^2 and 0.5mr^2*(omega)^2. Lowest if the mass is evenly distributed, higher if the mass is distributed more towards the edges.
Okay thank you for the answers i know it seemed like an obvious thing to me but i just wanted some reassurance. again thanks 