Physics of the Flywheel Launcher

VEX V5 Robotics Competition V5 Nothing But Net (15/16)
61

In the absence of air, the distance traveled by ball would be:

Where θ and v are the angle and speed at which the ball is launched. The g term is the acceleration value of earth’s gravity.

If you launch at 45 degrees, then that equation simplifies to:

Kinetic energy of the ball is determined by

Just make sure you do all your calculations in metric units.

If you make the bold assumption that the kinetic energy transferred to an NBN ball will be the same as the energy now being transferred to your softball, you can work out what your new distance will be.

However, all of the above ignores the effect air friction will have. The denser ball will “punch through” the air more effectively than the NBN ball (look up ballistic coefficient for more info on that). So it’s hard to say what will actually happen without calculating or testing the ballistic coefficient, and spin on the ball can also play an important role.

So, assuming a 45 degree launching angle and ignoring air friction and spin, your softball is traveling at:

V of softball = sqrt(d*g) = sqrt( (2.5 meters) * (9.8 meters/s^2) ) = sqrt( 24.5 meters^2/s^2)) = 4.95 meters/s

Using the equation for kinetic energy, we discover that your softball was launched with an energy of:

E of softball = 0.5*(0.206 kg)*(4.95 meters/s)^2 = 2.5 Joules.

So if a NBN ball were launched at 45 degrees and given an energy of 2.5 Joules, how far could it travel? We know the NBN balls have a mass of about 50 grams, which is 0.050 Kg.

V of NBN ball = sqrt( (Kinetic Energy) * 2/(Mass of NBN ball) ) …or…

V of NBN ball = sqrt( (2.5 Joules) * 2/(0.050 Kg)) = 10 meters/s

So the distance the NBN ball would travel is D = V^2/g = ((10 meters/s)^2)/(9.8 m/s^2) = 10.2 meters

But, in reality the air friction is going to trim that down substantially. The NBN balls are not very dense.

62

I’m fairly certain this wasn’t discussed, but let’s assume we have a single flywheel made of two wheels. How would the gap between the wheels affect the final velocity of the ball?

Someone please correct me if i’m wrong, but my logic would say that the final velocity of the ball would approach the tangential velocity of the wheel as the gap is widened and assuming there is still a 100% conversion of forces and no resistance.

This because we agreed that in a perfect 2d world, a single flywheel will shoot a ball with half the velocity of the flywheel’s tangential velocity. However, if we turned this 3d and made the gap between wheels exactly the same width as the ball and make perfect velocity vectors on both sides of the ball it should operate like a shooter with adjacent flywheels. We determined before that a flywheel with adjacent flywheels would travel at an equal velocity to the tangential velocity, so I would think that the transition between a perfectly centered flywheel to a wider and wider double flywheel would result in a higher final velocity.

63

So if I understand correctly, you want the ball to be squeezed between two wheels spinning in unison mounted on a single horizontal axis? That might work . . .
However, you might have problems with the balls getting stuck between the wheels and not releasing nicely

64

You could add a slight angle to the wheels as to funnel the balls between the wheels. But at that point you might as well have normal fly wheels.

65

I guess at the one extreme you would have the ball perfectly sandwiched between two wheels, the two wheels spinning synchronously, with the ball just barely touching each wheel at points just inside each wheel’s circumference. I suppose you could theoretically spin each of those wheels on a different axis so the wheels form a slight V shape, so the balls would be captured at one point, spun around at the tangential velocity, then released as the ball is moved around the orbit until the V opens up. But, in that case, I don’t see it being much different than a two-wheel flywheel shooter.

If the two wheels perfectly sandwich the ball, then I think you would have the problem Highwayman pointed out, in which the ball doesn’t have anything to cause it to release (except maybe centrifugal effects? or maybe some kind of kicker mechanism?).

If you try to roll the ball in some intermediate range (not entirely sandwiched but also not rolling along as you see in a single flywheel design), um, I’m not sure what you would get. Because you wouldn’t be rolling the ball against the opposite wall at the ball’s largest circumference… uh… you’d have to crank the numbers and see what really happens. I guess we never know until we try. However, whatever you do, I wouldn’t try to accelerate the ball by using the sharp “corner” of a wheel’s circumference: it will most likely just rip at the ball.

66

Great minds think alike! :slight_smile:

67

Putting anything other than a very slight V shape wheel axes seems wrong. It only allows momentary spin up and throws away the advantage of a single wheel shooter which is its long spin up roll. The sandwiched case is a good compromise to stabilize the roll. It still provides a normal force to hold the ball against the wall while providing a channel to stabilize the ball laterally. But doing this will reduce the w_wheel if one keeps the same exit speed. The best method is to have a single wheel on top with the channel formed into the wall. This method has an advantage because w_wheel must be higher to get the same exit speed which means higher energy flywheel.

The ball speed , V_ball, is determined by V_ball =w_ball r_ball and w_ball = w_wheelr_wheel/d where d is the distance above the wall where the wheel is tangent to the ball. So V_ball = w_wheelr_wheelr_ball/d . With the wheel on top of the ball, d = 2*r_ball , with it entirely sandwiched, d = r_ball. But in all cases it will impart spin to the ball which the two wheel balanced shooter will not. So the analog is not quite the same.

68

Yes, you are understanding correctly.

You’re probably right, I didn’t think it was plausible in a realistic scenario. The point I’m trying to get at is how would this affect the calculations of a flywheel under this condition, as well as the previous things that was discussed.

Ok, this answered my question. But that seems counterintuitive to me

If i’m understanding the final equation correctly after substitutions, it’s


Velocity of ball= ((Angular Velocity of wheel)(Radius of wheel)(Radius of the ball))/Distance the wheel is above the wall

Because radius of the ball is being multiplied, that would imply that as the wheel approaches the form of a single wheel, it would have a higher velocity. However we already agreed that adjacent flywheels (whether it’s side to side or top and bottom) has a higher exit velocity.

So in this case, wouldn’t the radius of the wheel be on the bottom/ everything should be divided by the radius?

If that is true, it doesn’t have the proper units, so I would assume the distance between the wall and wheel would be a multiplier/ on top (excuse my lack knowledge in proper mathematical/scientific terminology). Also, another thing that doesn’t make sense is if the distance is zero, where the wheel and wall are perfectly in line, that would imply that the overall velocity is zero, yet wouldn’t it have a perfect force conversion since the ball doesn’t need to bypass a “hump” as it goes over the wheel?

69

You have an understanding of the equation I wrote…however, the only thing that changes when you move the wheel is the parameter “d” which is the distance of the bottom of the wheel from the wall. It doesn’t make sense to consider d < r_ball since a vertical wheel cannot contact the ball below its midpoint. “d” is in the denominator and as you lower the wheel, d gets smaller and V_ball increases as expected. The units of the equation are fine.
When the wheel is on top, d = 2r_ball, so V_ball = w_wheelr_wheelr_ball/(2r_ball) = w_wheel*r_wheel/2 (= 1/2 the tangential speed of the wheel.)

70

I guess im misunderstanding what the wall is. Do you mean the ramp which leads to the flywheel, the hood which keeps the ball from flying out when it hits the flywheel, or something else?

71

Sorry if I was unclear. The wall is my simple definition of the surface that the ball rolls on while in contact with the wheel. Hood, cowling, circular containment surface, fender …all good.

72

…AH HA!!! EUREKA

Alright, now that makes so much more sense. Thanks a bunch:)

73

Up to now this thread was focusing on theory and some speculations, but now many teams already went through the multiple designs and have solid launchers on the working robots. So what are the practical lessons we have learned?

There is a number of recent threads where people share their successes with two-motor non-flywheel launchers. There is also a number of threads, where (mostly younger) teams share their painful experiences with the flywheels. It doesn’t have to be that way.

I think continuous flow flywheel is a superior type of launcher. Experienced teams could build award-winning non-flywheel launchers that will easily beat an average flywheel, but if they put their skills into perfecting the flywheel design, I think, they could do even better.

Most of the problems that people are experiencing with the flywheels are due to the excessive friction, too much of the ball compression, and even more friction as a result of that.

Here is our attempt of addressing these problems by building an extremely low-friction flywheel prototype:

Earlier vamfun presented a convincing argument that a single flywheel with large moment of inertia, running at twice the speed of the dual flywheels, is an optimal solution. That is, if you can successfully manage the friction.

The purpose of this video is to show that two 393 motors provide enough power for a single flywheel launcher. There are two turbo motors driving flywheel with 19.2:1 external (46:1 total) gear ratio. Not only there is enough power to reach the high goal, but flywheel rotates way too fast and keeps overshooting. We will fix that later.

I haven’t done precise measurements, but from the crude video analysis it looks like if you feed the balls at the right speed you could consistently hit target while shooting 4 balls in about 2 seconds. If you feed balls as fast as you can - it will launch four balls in just over a second with the first three overshooting, and the last one falling slightly below the target. Clearly at that speed flywheel has enough kinetic energy stored to give the first three balls more than enough speed. Obviously, for this design to be practical the feed rate needs to be program-controlled like 323Z does.

Single flywheel concept also fits nicely with reducing robot’s weight, size, and the number of the moving parts. The last is especially important, because it will let you reduce the number of axles.

Technik jr started building our first prototype with 21:1 total ratio driven by two force motors. It could launch balls only a few feet away. It would not even spin up with one force motor. Getting down to one force motor was relatively easy with just better build procedures, but moving further was much harder.

Power loss at each axle is proportional to (torque due to friction) * (angular velocity). Therefore even 0.5 in-lbs in resistance at the flywheel bearings would amplify 21 times to 10.5 in-lbs around the motor axle ( out of maximum available ~15 in-lbs stall torque). However, you want to run motors at only ~10% of the max load for the best efficiency, spinning at about ~85% of their idle speed (see the graph in the post below).

According to our estimates for a single 5" flywheel that would require ~ 30:1 total gearing. Since each axle introduces friction, you want to reduce total number of axles. Both speed and turbo motors could be used with just three axles and require ~19:1 (speed) and 12.5:1 (turbo) external gear ratio. I think turbo is a better choice because smaller torques will flow through the system.

So the next task was to actually reduce the friction so much that the launcher could be spun up with a single speed and, eventually, turbo motor. Technik jr and teammates have developed better alignment procedures and it let us, at least, start with a single speed motor, but launching balls or starting up with a single turbo motor was still unreachable. There was just too much friction between square axle and the bearings, which would only increase when flywheel would try to compress the ball.

So we did two things: first we introduced a pre-boost stage to pre-compress the balls at the slower speed, when more torque is available, and second, and more important, we figured out how to build a better bearing. I still cannot reveal the secret of how team did it, but I could give you a hint that they found an almost a perfect VRC-legal part for this mission, broke 1/8, and stripped 13/64 drill bit while perfecting the implementation :slight_smile:

74

And continuing discussion from another thread:

I think, your gear ratio is way too high. Since product of (torque)*(angular velocity) is limited by the motor’s power you don’t have enough torque to overcome the friction. Try reducing gear ratio to ~30:1 total and see what happens.

I just disconnected wires from one of our motors, and the launcher was still able to spin up and hit the target even with the dead weight of the second motor on its back. The recovery time is about 6-7 seconds, but it doesn’t stall.

As for the number of 5" flywheels, affecting the moment of inertia. We use about 1.7 - 1.8 worth of 5" flywheels in the final stage (yes that’s what I meant) and it looks like overkill at those speeds. We will reduce our gear ratio down to 30:1 first and then see if it needs to be adjusted further. Heavier (or more) flywheels shouldn’t stall the motors - it will just increase spin-up time. It must be friction that is stalling the motors.

Finally, I don’t know the right answer about flywheel sizes for the intermediate stage and how it affects ball pre-compression. We still have to experiment with the shape of the backplate and will, probably, be tweaking it until the day of the competition.

75

I haven’t seen your design, but it would depend on how fast your flywheel is actually spinning. If you have too much friction in the flywheel bearings then you are pushing your motors into the high load, slow speed area (to the right on the graph).

You, really, want to be on the left, around 1.5 in-lbs torque and 85% nominal speed while launching the balls. There could be excursions into above 90% speeds when idling and below 85% when motors are recovering flywheel speed. The more kinetic energy flywheel could store the narrower that band could be. However, you don’t want flywheel to be too heavy because, at some point, it won’t be practical and it will take forever to bring it up to speed.

Here are jpearman’s torque-speed curve for 393 motor:

So your 8:1 and 32:1 ratios look about right and, if the motors rotate at ~85% of their nominal speed, then you want to look at the “good” friction. In order for flywheel to transfer some of its kinetic energy and angular momentum to the ball, there needs to be a good grip between them. That’s why we put rubber bands there and spent extra time making sure they are evenly distributed.

Also, there has to be just the right amount of ball compression. If compression is too small, grip will be weak and the balls will not fly far, even if the flywheel rotates fast enough. If the compression is too large then a lot of energy will be wasted compressing the balls.

Michael, I am sure you know this already, because I’ve seen 6210z robot’s reveal and it handles balls pretty well. My main concern would be how to compress balls of varying densities. We haven’t experimented with backplate design much yet - in the video we we just ramming the balls through with the excessive force.

But the next area of our research would be to try to design backplate in a such way that a softer material will handle a lot of the compression and, therefore, provide more predictable compression and ball-to-flywheel gripping force, regardless of the ball density.

76

Thank you for all of your help and advice. Wednesday I will be able to implement all that you have told me and I will let you know if we find a backplate design that handles compression well.

77

My team had a meeting today, and decided to conduct an experiment. They replaced both turbo motors with the speed motors and fired up the launcher.

Even though the launcher has 30:1 total and 19:1 external gear ratio, which I claimed to be perfect for the speed motors, the balls, embarrassingly, never made it to the target. At the best shooting angle they were crossing 3ft height on the way down at least 2 ft short of the high goal.

Lets pretend that I am a rookie team that trusted advice from this thread to build the launcher with 30:1 ratio. So I came back to this thread looking for help. Now, putting back on my technik3k hat I have advised myself to do some debugging. In particular, I told myself to report back the angular velocity at which flywheel is rotating.

Fortunately, we have a piece of code to display (in the debug / globals tab) the battery voltage and quad velocity estimate (in rotations per second) :

float vel=0; 
short vBat=0;
float mPow=0;

task main()
{
  while(1)
  {
    long t0 = nSysTime;
    short q0 = SensorValue[Quad1];
    vBat = nImmediateBatteryLevel;
    mPow = (3*mPow+SensorValue[Button1])/4; // when button is pressed ramp up motor power gradually
    motor[M1]=127*mPow;
    motor[M2]=127*mPow;
    delay(50);
    vel = (3*vel + ((SensorValue[Quad1]-q0)/0.36)/(nSysTime - t0))/4; // smooth velocity estimate
  }
}

So I have run several test cases noting motor speed (with percentage of the nominal) and battery voltage.

No motors are running: vBat = 7.86
Motor 1 idle: 2.78 rps (104%) v=7.80
Motor 2 idle: 2.61 rps (97.8%) v=7.78
Motor 1 driving flywheel alone: 2.28 rps (85.5%) v=7.56, balls are about 4 ft short of target.
Motor 2 driving flywheel alone: 2.05 rps (76.8%) v=7.46, balls are about 5 ft short of target.
Both motors are driving flywheel: 2.51 rps (94.1%) v=7.55, balls are about 2 ft short of target.

So what can you tell me, beside that I need to increase my gear ratio?
Is there any guesses as why my 30:1 prediction fell short?
What do I need to work on, if I want to perfect my design?
What physical phenomena do you think is responsible for this shortfall?
Is there any other interesting information you can deduce from this data?

Please, feel free to share any thoughts or ask additional questions. This should help everybody in debugging their flywheels.

Also, there will be personal prizes in the form of sharing top-secret stuff from our designs with the people who gives correct answers to the above questions. There is a limited number of valuable secrets, so hurry up to get exclusives. :slight_smile:

78

I first want to mention that it seems that your motors are not optimally matched. clearly motor 1 is more efficient and powerful as seen by the numbers. Motor 2 also is taking a bigger draw from the battery. so it may be a good idea to switch it out and see if that helps in any way

79

Here is our flywheel after we redesigned it today. We got it shooting at around 80% accuracy with a 13:1 and 39:1 gear ratio but it stalled after about 7 or 8 shots. After this we reduced the ratios to 13:1 and 26:1 and it didn’t stall but the balls were landing a foot below the lip of the goal when positioned inside the loading zone. My best guess is that the optimum gear ratios are somewhere a little above 10:1 for the first wheel and somewhere around 30:1 for the second wheel. I am hoping to test out some of the following ratios next week: 13:1 and 32.5:1, 11.2 and 28:1, and 13:1 and 26:1; that last one being built slightly different (positioned vertically with a curved back, sort of like 323z’s).

IMG_1219.JPG
IMG_1219.JPG2592×1936 449 KB

80

My best guess is that your compression is not working well enough. We put a C-channel across the top of our entire system and 30:1 seems to be the best gear ratio for that design (although I am not sure whether or not that compression technique hinders our rate of fire).