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Unread 11-16-2001, 11:53 AM
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Question about how Drive Train push the robot... shouldn't the force accelerate the robot?

After thinking about motors... I started thinking about how a drive train move the robot.

As far as I can tell... The motors used in the drive train produce a torque and angular velocity on the gears and wheel, and the wheel transform those two into linear force and linear velocity parallel to the ground.

Now, the linear force from the wheel is limited by friction, which is F = u (mu) x N (weight on powered wheels), nevertheless, there is a force from the wheel pushing against the ground. And in return, the opposite force pushes the robot forward.

Now, I understand that the speed of the robot (linear speed) can be calculated with wheel RPM and radius. But, how do the force come into play?

When the robot accelerate from 0 velocity to the linear velocity, the robot have a resisting force F = m X a, and that robot will start moving only when the linear force from the wheel got bigger than that resisting force.

Well, then... when the robot start moving, that mean the wheel force is in fact larger than resisting force of robot... and there is some extra force left, which is bigger with a more powerful motor.

I know that the left over force accelerate the robot until it reach the max speed. At that point, the robot stop accelerating. After all, the robot can't possibly go faster than the speed generated by how fast the wheel spin.

So, my question is, if there is still a force acting on the robot from the wheel, and that F = ma, then how come the robot isn't accelerating due to that force? I don't suppose drag got so big that it cancel out that pushing force... So, where did it go?

Is this left over force (wheel force - resisting force) the same as the force when robot is pushing against something (i.e. another robot)?

Do this left over force only appear when something external is pushing against it?

Or is it because the fact that we are talking about "power", and that the force is no longer a factor of velocity of the object?
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Unread 11-16-2001, 08:33 PM
Jay Lundy Jay Lundy is offline
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I'm only taking my first year of physics this year and I don't know much about motors but I'll give it a try.

First of all, I don't think the friction between the wheel and the ground has much to do with it. However, it becomes important when talking about traction, and at what point the wheels will start slipping. Friction is the only thing that allows for the transfer between rotational and translational movement. For example, if you ran your robot on a frictionless surface, the wheels would just spin and your robot would not move. Conversely, if you gave your robot a push on a frictionless surface, the wheels would not spin, but the robot would move.

Relating this to last year's game, the bridge was at an angle so N was less, so the force of friction was less, so the wheels slipped. Basically traction depends on the coeffecient of static friction between your wheels and the floor, your weight, and the angle of the bridge. If the force causing your wheels to rotate overcomes the coeffecient of static friction * N, then the wheels slip.

As for why the robot does not keep accelerating, I seem to remember seeing a chart that graphed torque vs. speed. As the speed got faster, the torque got lower. Eventually, there is no more torque and since T=Fd, there is no more force to cause the robot to accelerate.

Think of it like you trying to rotate something like a bike wheel. You can make it accelerate, but eventually the speed maxes out and you can't apply any force to make it go faster. I think the motor works similarly.

Basically, the only significant forces working against the robot is the friction in the wheel axle, and any other objects the robot may run in to. Although the coeffecient of friction in the ball bearing may seem small, it makes a difference when it has 130 lbs on it. Air resistance and the friction between the wheel and the floor are both very small and don't affect the robot much.

Of course, I may be wrong about any of this.

Last edited by Jay Lundy : 11-16-2001 at 08:44 PM.
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Unread 11-16-2001, 10:25 PM
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Quote:
Originally posted by Jay Lundy
I'm only taking my first year of physics this year and I don't know much about motors but I'll give it a try.

First of all, I don't think the friction between the wheel and the ground has much to do with it. However, it becomes important when talking about traction, and at what point the wheels will start slipping. Friction is the only thing that allows for the transfer between rotational and translational movement.

Relating this to last year's game, the bridge was at an angle so N was less, so the force of friction was less, so the wheels slipped. Basically traction depends on the coeffecient of static friction between your wheels and the floor, your weight, and the angle of the bridge. If the force causing your wheels to rotate overcomes the coeffecient of static friction * N, then the wheels slip.

As for why the robot does not keep accelerating, I seem to remember seeing a chart that graphed torque vs. speed. As the speed got faster, the torque got lower. Eventually, there is no more torque and since T=Fd, there is no more force to cause the robot to accelerate.

Think of it like you trying to rotate something like a bike wheel. You can make it accelerate, but eventually the speed maxes out and you can't apply any force to make it go faster. I think the motor works similarly.

Basically, the only significant forces working against the robot is the friction in the wheel axle, and any other objects the robot may run in to. Although the coeffecient of friction in the ball bearing may seem small, it makes a difference when it has 130 lbs on it. Air resistance and the friction between the wheel and the floor are both very small and don't affect the robot much.

Of course, I may be wrong about any of this.

You are actually right in a lot of parts. I really said it wrong when I said the linear force on the wheel is limited by friction. What that really means is that the linear force coming from the wheel can only be so strong before it over come friction and start slipping...

And, you are right on with the situation on the bridge. When even a robot gets on the bridge, there's a certain force it need to push up the bridge.

HOWEVER, the speed torque curve is used to show the motor's responds to torque. When there's no load on the motor, it spin with a speed we call free speed. Then, as we put more and more load onto the motor, it spins slower and slower, until the motor stall. The amount of torque it takes to stall a motor is stall torque. With those two numbers, we can get a speed torque curve. So, the amount of torque didn't got less because motor speed is higher... but the other way around.

You see, when you mention about how rotating a free wheel, you are right that it can only go up to a certain speed. That happen because the force you use to spin the wheel is much less in the reference frame of the wheel. Imagine if you are using your hand to spin a bike wheel. When the bike wheel spin up to a certain speed, your hand simply isn't fast enough to catch up with the wheel and exert the same amount of force on the wheel.

So, when that happen to the drive train, as the robot goes faster and faster, the load on the motors is no longer as much as before, and therefore it should be spinning faster. So the RPM reflecting back to the gear driven by motor will catch up to the motor's RPM, and forces from the motor is only used to over come friction and other forces.

Hmm... Is this true? Is the speed of the motor maximum when robot is accelerated enough such that the momentum of the robot is carrying the robot forward and the load on the motors got much less than not zero speed?

Well, my question still remains. How do I calculate how strong the robot is pushing under different situation? How should we set up gear ratio when the force/load is changing all the time? And how do we calculate how much force it takes just to move the robot across the field, even when not going up the bridge? If the robot is dragging heavy objects such as two goals... how do we calculate the force it takes to push two goal + robot?
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Unread 11-16-2001, 11:51 PM
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Simply Friction

The electric motors exert a constant force relatively speaking on the system.

Basically, as the speed of the robot increases, the internal friction (friction in the gears/sprokets/bearing) increases directly proportional to the increase in speed.

So basically if comes down to this.

(Force from motor)-Friction (energy lost to heat/sound) = remaining force left to accelerate your robot.

Remember the force from the motor is constant, but the force of friction increases with increasing speed. So as you accelerate there is less and less energy in reserve left to accelerate your robot. At some point, the

force of friction = force put out by motors.

at this point, the robot has reached a dynamic equilbrium and can no longer accelerate because all of the force is being used to overcome the fraction in your system.

If you think about it, the drill motor drive train is relatively noisy, (loss of energy) and the drill motor heats up considerabally after use (loss of energy).
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Unread 11-18-2001, 01:43 PM
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Re: Simply Friction

Quote:
The electric motors exert a constant force relatively speaking on the system.

Basically, as the speed of the robot increases, the internal friction (friction in the gears/sprokets/bearing) increases directly proportional to the increase in speed.

This is more or less backwords.

The losses in the system will not actually increase that much as the robot accelerates.

The force (torque) output from the motors will decrease as the robot accelerates.

There is a limited amount of power available from the motors. The power from the motors is equal to the product of the torque they provide and the speed at which they are turning. As the RPM's of the motors go up, the amount of torque they can supply goes down. Eventually, the available torque is reduced to the point where it is equal to the frictional losses in the system and the robot stops accelerating and continues on at constant speed.
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Unread 11-18-2001, 02:33 PM
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My bad.

I completely forgot about that chracteristic in electric motors.

Too much thinking in the "ideal" physical world is not healthy, I need to start working with robots again instead of textbooks.
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Unread 11-18-2001, 02:38 PM
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Re: Re: Simply Friction

Quote:
Originally posted by P.J. Baker


This is more or less backwords.

The losses in the system will not actually increase that much as the robot accelerates.

The force (torque) output from the motors will decrease as the robot accelerates.

There is a limited amount of power available from the motors. The power from the motors is equal to the product of the torque they provide and the speed at which they are turning. As the RPM's of the motors go up, the amount of torque they can supply goes down. Eventually, the available torque is reduced to the point where it is equal to the frictional losses in the system and the robot stops accelerating and continues on at constant speed.
So, the only way to measure the robot speed is to either measure the friction in the system... or to actually record the speed of the robot (like timing the robot's time to race across the field". I don't suppose there's any other way to do this.

As for the force the robot is pushing, since the frictional force from the wheel to the ground is the maximum amount before the wheel start the skip, we can only increase the mass of robot or better traction on the wheels to increase how hard it's pushing.

When a robot is pushing another around, what's really happening is that the robot have more traction/mass/ratio/power, where traction and mass increase the max. frictional force on the wheel, and a better ratio to keep the motor from stalling/heating up (which reduce motor power and draw too much current).

And of course, making sure the wheels are always touching the ground is important too.

May I assume that the Technokats robot from last year is going to have the strongest drive train to push everyone around since they have extremely good traction fron tank tread, and dual motors design with 4 motors on the drive train?

Last edited by Ken Leung : 11-18-2001 at 08:14 PM.
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Unread 11-18-2001, 09:18 PM
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In my observations...

I have never seen a twin drill motor drive stall when the drill motor is set in low gear and running anywhere between 2:1 or 3:1 gear ratio with an 8 inch wheel.

Usually the case is, when two robots are pushing against each other, the frictional force between the wheel and ground is overcome before the stall torque of the motor.

If you're running in high gear on the drill motor, that's another story. Generally, (especially when you're trying to tow two goals up to balance) the motor stalls when you begin to climb the ramp.

I don't know how exactly would be the best way to find the best combination of frictional force and weight. With a lighter robot, you can accelerate and acheive a higher top speed. On the same note, you do not have as much mass to push other robots around with and the frictional force when you are climbing a ramp or bridge is less than it could be.

The design challenge for us never came to this point however, we were always 130.0 lb. (haha)
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Unread 11-18-2001, 09:47 PM
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Stalled motors

The first symtom you encounter if your robot is geared too high is that the 30 amp breakers will intermittently kick out. The drill motors draw LOTS OF CURRENT, like 100 amps or so, when stalled. You will not stall the drill motor longer than a second or two before the breakers trip. When you are geared "bordering on too high," the motors will run very hot and smell like they are burning up which, come to think of it, they are.

Our 2001 robot with tank drive and gear change is geared about as high as will work reliably in high gear under non-pushing and non-climbing conditions. In low gear, even with very good traction, it will slip rather than tripping breakers or stalling motors. We can run in low gear continuously for hours without the motors getting too hot. With the motors lightly loaded, the little fans inside do their job very well.
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Unread 11-18-2001, 10:24 PM
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drawing too much currents...

Pretty much the reason those drill motors draw so much current is because we put too much load on them. That is why when robot go up the bridge with goals or do 360 turn with high traction tank drive... And at the same time, those motors get really inefficient with high load, losing a lot of motor power to heat.

So, we fix that with either a bigger gear ratio, and/or more motors. And I think more motors is the way to go since bigger ratio comes with a slower speed.

By the way, Teams always had to consider between weight and speed when they design their robots. But a lot of competitive robots get really close to weight limits anyway due to other components. So we can pretty much assume that during the competition we will face some heavy 130lb robots. However, robots can have some good advantages with light weight and fast drive train. That is why there were many interesting combinations of robot during 1999 and 2000 game...

And, If we are going head to head competition, which I think is extremely likely, robots will need some serious pushing force if they plan to push others around... no matter they are defensing or over-coming opponents' defensive robots.

Come to think of it, the best ways to push against other robot is to somehow reduce their traction... or have more torque on the wheels... but of course I won't be telling how exactly to accomplish those conditions just yet
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Unread 11-21-2001, 12:02 PM
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Variable Gear Ratio

You know there is another solution .... Dynamically varying gear ratio. The TechnoKats and others have designed the switching gearbox. There is a way to change gear ratio dynamically (without stoping and shifting). All I will say is:

2 motors + 1 planetary gear set = Speed ratio from 0 to 10:1

How can you get 0? That is easy ... both motors spinning but your output shaft is sitting still creating tremendous amounts of torque. Sound impossible? Well, it's not and we will have video to prove it by January. Stay tuned ....


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Unread 11-21-2001, 01:49 PM
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Re: Variable Gear Ratio

Quote:
Originally posted by Paul Copioli
2 motors + 1 planetary gear set = Speed ratio from 0 to 10:1
ChrisH has had in mind to do just that for Beach Bot the last two years, but for various reasons it hasn't worked out. If anyone succeeds in implementing this, I (and he, I'm sure) would really like to know.
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Unread 11-26-2001, 09:39 AM
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Ken has discovered my secret idea for the 2002 season, so I will reveal it to you all at this time. There is extra energy lost to friction, mostly by friction that is mostly transferred into heat mostly in the motors for a while. Now the secret is to hook up an ideal heat engine between these warm motors and a giant tub of liquid nitrogen in your robot (or another source of relatively low temperature). This should create an additional amount of energy you can feed back to your speed controllor in series and you will have somewhere in the range of 16.1-17.4 V for full force. This in fact creates even more heat in your motor which in turn gives you even more voltage, so you'll get a lot more speed or torque depending on your religion.

Also, someone mentioned sound as a way energy is lost. This is a major problem since tons of energy is lost this way. After careful consideration I believe the best solution is to muffle your robot. If the sound can't get out, you can't lose the energy!

Patrick
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