### Wednesday, June 15, 2005

## Polymorph Carriage Tests

The first tests have been done on the Polymorph carriage, oiled lightly, with the feet trimmed to 75% of the circumference and de-stressed with a hair dryer. Smoothly overcoming the friction between the four saddles requires a force of 2.6N and 1.2N keeps it moving:

The carriage itself weighs 140g and I started to wonder just how much power we're going to want to expend on moving the carriage around at this point?

As an example, the tiny little 3V-6V Meccano motor I have here consumes 200mA startup current at 3V and will move the carriage. Assuming that's around 50% efficient, thanks to the gearbox, and a load on the carriage that is expected to be about 300g, the dilligent student should be able to work out the maximum speed and acceleration. It's been a good few years since I did physics at school...

Would anyone like to tell me:

a) The likely top speed.

b) What's the accceleration, and

c) Is this good enough?

Vik :v)

PS Spot the cameo appearance of the FDM'd pinch wheel assembly.

The carriage itself weighs 140g and I started to wonder just how much power we're going to want to expend on moving the carriage around at this point?

As an example, the tiny little 3V-6V Meccano motor I have here consumes 200mA startup current at 3V and will move the carriage. Assuming that's around 50% efficient, thanks to the gearbox, and a load on the carriage that is expected to be about 300g, the dilligent student should be able to work out the maximum speed and acceleration. It's been a good few years since I did physics at school...

Would anyone like to tell me:

a) The likely top speed.

b) What's the accceleration, and

c) Is this good enough?

Vik :v)

PS Spot the cameo appearance of the FDM'd pinch wheel assembly.

Comments:

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Isn't it more of a power=force*distance/time

therefore power/force=distance/time so

.3 W/1.2 N=.25 m/s? I may be wrong - haven't been doing teriffically in my physics classes - but I think that's it.

I do remember that acceleration is a somewhat harder thing to get from power, if only a bit. More tanlged.

Hmmm... one meter every four seconds? certainly not tremendously fast, but I can see that being fast enough for basic operation.

therefore power/force=distance/time so

.3 W/1.2 N=.25 m/s? I may be wrong - haven't been doing teriffically in my physics classes - but I think that's it.

I do remember that acceleration is a somewhat harder thing to get from power, if only a bit. More tanlged.

Hmmm... one meter every four seconds? certainly not tremendously fast, but I can see that being fast enough for basic operation.

I think some practical experiments are going to be called for here.

Remember that this is the smallest motor I've got - I'm trying to establish what the requirements are at this point, and I can add more horsepower simply by running at 6V.

Vik :v)

Remember that this is the smallest motor I've got - I'm trying to establish what the requirements are at this point, and I can add more horsepower simply by running at 6V.

Vik :v)

Just a note Vik,

I agree that Power = Force * distance/time However, I think your using the wrong force in the equation. The forces you measured are the Static Frictional Force (2.6N) and the Kinetic Frictional Force (1.2N). These forces are opposing forces to the force moving the carriage. The force moving the carriage is due to the motor, that is the force which will yeild the max velocity and accelleration. The motor will be required to overcome the frictional forces.

As a side note, the coefficiants of frictional force you measured would be as follows:

For static friction:

2.6N/(.140kg*9.8m/s^2) = 1.9

For kinetic friction:

1.2N/(.140kg*9.8m/s^2) = .875

Which seem like pretty high frictional forces by the way. A block of rubber sliding on pavement has a kinetic frictional force coefficient of about 1.

I agree that Power = Force * distance/time However, I think your using the wrong force in the equation. The forces you measured are the Static Frictional Force (2.6N) and the Kinetic Frictional Force (1.2N). These forces are opposing forces to the force moving the carriage. The force moving the carriage is due to the motor, that is the force which will yeild the max velocity and accelleration. The motor will be required to overcome the frictional forces.

As a side note, the coefficiants of frictional force you measured would be as follows:

For static friction:

2.6N/(.140kg*9.8m/s^2) = 1.9

For kinetic friction:

1.2N/(.140kg*9.8m/s^2) = .875

Which seem like pretty high frictional forces by the way. A block of rubber sliding on pavement has a kinetic frictional force coefficient of about 1.

Thanks for putting me straight on that. Like I said, long time since I did physics and I was never a star pupil at it :)

The friction is higher than I'd like, but the thing needs to grip the rails well enough so that it doesn't wobble about. It's not just resting on the top of the rails, but wrapped around them and hanging on so it's not just a downward-acting force generating the friction.

Vik :v)

The friction is higher than I'd like, but the thing needs to grip the rails well enough so that it doesn't wobble about. It's not just resting on the top of the rails, but wrapped around them and hanging on so it's not just a downward-acting force generating the friction.

Vik :v)

In the "Yet another polymorph extruder", someone mentioned a flow rate figure of 0.4 mm3/sec. It seems to me that at such slow deposition rates, the speed of the carriage is the least of your problems.

Hence, I would expect you could just gear the motor *WAY* down to get the power to move it without having to be concerned at a lack of speed...the limitations of flow rates at the nozzle being a far bigger concern.

When I've built positioning mechanisms like this in the past (using Lego Technics components) I've found that using a rack and pinion arrangement works better than something that slides on rods or whatever.

The advantages are twofold:

1) You aren't dependent on high frictional forces to avoid slippage on the track.

2) You can know your position precisely just by counting the number of revolutions of the motor - which you can do with a simple optical shaft encoder - such as you can find in a computer mouse.

You ought to be able to fabricate a pretty decent track using the RP machine itself.

Hence, I would expect you could just gear the motor *WAY* down to get the power to move it without having to be concerned at a lack of speed...the limitations of flow rates at the nozzle being a far bigger concern.

When I've built positioning mechanisms like this in the past (using Lego Technics components) I've found that using a rack and pinion arrangement works better than something that slides on rods or whatever.

The advantages are twofold:

1) You aren't dependent on high frictional forces to avoid slippage on the track.

2) You can know your position precisely just by counting the number of revolutions of the motor - which you can do with a simple optical shaft encoder - such as you can find in a computer mouse.

You ought to be able to fabricate a pretty decent track using the RP machine itself.

A belt drive is just as stable - most printers use this technique. I'll be experimenting with one this weekend, and Ed successfully used a belt drive on his axis prototype.

What positional accuracy did you get out of the Lego?

Vik :v)

What positional accuracy did you get out of the Lego?

Vik :v)

Positional accuracy of a Lego system rack and pinion is probably around 1mm - which clearly isn't good enough for the RepRap machine.

The reason Lego does so poorly (I believe) is because the pitch of the teeth on the rack parts is so large. A finer pitch would help out a lot.

Also there is quite a lot of slop in the gear teeth because the same Lego gear wheels must be used for meshing gear wheels together as are used for meshing into a rack. Ideally, one would want teeth shaped specificially for the purpose - but the Lego tooth shape is an ugly compromise.

The reason Lego does so poorly (I believe) is because the pitch of the teeth on the rack parts is so large. A finer pitch would help out a lot.

Also there is quite a lot of slop in the gear teeth because the same Lego gear wheels must be used for meshing gear wheels together as are used for meshing into a rack. Ideally, one would want teeth shaped specificially for the purpose - but the Lego tooth shape is an ugly compromise.

For comparison, the width of a tooth on a printer drive belt is usually in the order of 1.8mm

What is the Lego like?

Vik :v)

What is the Lego like?

Vik :v)

To make the most 'replicatable' parts possible, could we look at using the machine to produce 'vee and flat' style slides, driven by rack and pinion?

At least all the parts would be able to be self-replicating...

At least all the parts would be able to be self-replicating...

I understand that the slider fit must be tight for accuracy reasons, but I think the high friction will be a limiting factor for many reasons. Static friction could become a major problem for starting and stopping accurately. Ideally the static and kinetic coeffecients should be very close in value for smooth accurate starting. Heat buildup due to friction could also cause the polymorph to fatigue and loss of tollerance could result. What about a brass collar around the shaft. I'm sure you could find an "off the shelf" brass bushing with a good fit. Or better yet a teflon bushing. Teflon on steel has a VERY low Static and Kinetic coefficient of friction (about .04 for both).

A simple test to find or compare frictional coefficient is to simply tilt the rod and wait for the carriage to slide. If the rod can be tilted to vertical, the coefficient is greater than one, and probably too high. If the cariage slides at 45 degrees the coefficient is .5 (It's a force vector thing) 22.5 degrees means .25 etc...

From the coefficient it's simple to find the force required. It's simply

Force to move = frictional coefficient * mass of object * Gravity

Here's a site with some coefficient of friction data:

http://www.physlink.com/Reference/FrictionCoefficients.cfm

I'd also like to note that the resolution of a rack and pinion is not due to gear pitch. It's due to two things:

1. Gear tolerance. Slop between the gears will cause loss of accurracy.

2. Your ability to start and stop the motor precisely. If you were to use a stepper motor, for example, the resolution would be the detents, or number of steps, the motor was cabable of.

Two high tolerance gears have infinite posisions regardless of gear pitch (as do cog and belt systems). That's why your printer can print with a resolution VASTLY higher than 1.8mm even though the cog and belt system it uses has a 1.8mm pitch.

-Brian

A simple test to find or compare frictional coefficient is to simply tilt the rod and wait for the carriage to slide. If the rod can be tilted to vertical, the coefficient is greater than one, and probably too high. If the cariage slides at 45 degrees the coefficient is .5 (It's a force vector thing) 22.5 degrees means .25 etc...

From the coefficient it's simple to find the force required. It's simply

Force to move = frictional coefficient * mass of object * Gravity

Here's a site with some coefficient of friction data:

http://www.physlink.com/Reference/FrictionCoefficients.cfm

I'd also like to note that the resolution of a rack and pinion is not due to gear pitch. It's due to two things:

1. Gear tolerance. Slop between the gears will cause loss of accurracy.

2. Your ability to start and stop the motor precisely. If you were to use a stepper motor, for example, the resolution would be the detents, or number of steps, the motor was cabable of.

Two high tolerance gears have infinite posisions regardless of gear pitch (as do cog and belt systems). That's why your printer can print with a resolution VASTLY higher than 1.8mm even though the cog and belt system it uses has a 1.8mm pitch.

-Brian

Incorrect.

Suppose we have a surface which is raised an angle t from horizontal. Place a block on this surface. Assume that the coefficient of friction between the two is u. The weight of the block will be a vertical vector, which we can divide into the sum of two vectors, one parallel and one perpendicular to the surface. Call the parallel one Wx, and the perpendicular one Wy. The normal force will be equal in magnitude to Wy, and will be directed perpendicularly to the surface (in the opposite direction from Wy). So we say Wy=N. The frictional force, f, will be f = u * N = u * Wy, and will be directed in the opposite direction from Wx (parallel to the surface). A bit of quick trig will show us that Wy = W * cos(t), and Wx = W * sin(t). So, we have f = u * W * cos(t). Now, let t be the angle at which the block starts sliding. So, the frictional force will be equal to Wx. f = Wx = W * sin(t). Now we have u * W * cos(t) = f = W * sin(t). Divide out the W, we get u * cos(t) = sin(t). Divide by cos(t), and we get u = sin(t)/cos(t) = tan(t).

Thus, the coefficient of (static) friction is equal to the tangent of the angle at which the block starts sliding. If you can get the surface to vertical before it starts sliding, then your coefficient of friction is infinite. If you get it to 45 degrees, the coefficient is 1. 30 degrees -> 1/sqrt(3). etc.

Now if you want the coefficient of sliding friction, raise the surface to get it sliding, then lower it to the point where it stops sliding again. Measure that angle and take the tangent.

Of course, as Vic pointed out, the actual carriage wraps around the rod, which complicates things, and this simple "raise the plank the block is on and measure the angle" test doesn't work. In that case, actually measuring the coefficient of friction is quite a bit more difficult, but the basic formula is u = f * N, that is, coefficient of friction equals frictional force times the total normal force (force perpendicular to the surface per unit area, integrated over all points of contact).

Suppose we have a surface which is raised an angle t from horizontal. Place a block on this surface. Assume that the coefficient of friction between the two is u. The weight of the block will be a vertical vector, which we can divide into the sum of two vectors, one parallel and one perpendicular to the surface. Call the parallel one Wx, and the perpendicular one Wy. The normal force will be equal in magnitude to Wy, and will be directed perpendicularly to the surface (in the opposite direction from Wy). So we say Wy=N. The frictional force, f, will be f = u * N = u * Wy, and will be directed in the opposite direction from Wx (parallel to the surface). A bit of quick trig will show us that Wy = W * cos(t), and Wx = W * sin(t). So, we have f = u * W * cos(t). Now, let t be the angle at which the block starts sliding. So, the frictional force will be equal to Wx. f = Wx = W * sin(t). Now we have u * W * cos(t) = f = W * sin(t). Divide out the W, we get u * cos(t) = sin(t). Divide by cos(t), and we get u = sin(t)/cos(t) = tan(t).

Thus, the coefficient of (static) friction is equal to the tangent of the angle at which the block starts sliding. If you can get the surface to vertical before it starts sliding, then your coefficient of friction is infinite. If you get it to 45 degrees, the coefficient is 1. 30 degrees -> 1/sqrt(3). etc.

Now if you want the coefficient of sliding friction, raise the surface to get it sliding, then lower it to the point where it stops sliding again. Measure that angle and take the tangent.

Of course, as Vic pointed out, the actual carriage wraps around the rod, which complicates things, and this simple "raise the plank the block is on and measure the angle" test doesn't work. In that case, actually measuring the coefficient of friction is quite a bit more difficult, but the basic formula is u = f * N, that is, coefficient of friction equals frictional force times the total normal force (force perpendicular to the surface per unit area, integrated over all points of contact).

jonored's original post is reasonably accurate. P = m*a*d/t, m*a = F. So, a .3W motor will be able to deliver 1.2N of force over a distance of .25 m every second. That gives a top speed, assuming a frictional force of 1.2 N and 50% efficiency and all that, of .25 m/s as he said. That will decrease as load is added to the carriage, but how it will decrease is not easy to predict, since as I pointed out, a carriage is not a block on a board, and so the friction = coef * weight doesn't quite work here.

As for acceleration: I made some calculations, and I think that the power delivered by the motor must be a function of the velocity, because if it is constant, then the force applied at zero velocity is infinite. So, unless I'm doing something wrong, I think more information about the motor is needed.

I suspect that the best thing to do is just to simply try it out and make some measurements.

As for acceleration: I made some calculations, and I think that the power delivered by the motor must be a function of the velocity, because if it is constant, then the force applied at zero velocity is infinite. So, unless I'm doing something wrong, I think more information about the motor is needed.

I suspect that the best thing to do is just to simply try it out and make some measurements.

All this, of course, fails to account for any friction in the drive system, which is going to be an old HP DC motor and rotation encoder out of an ancient printer. Mostly on account of that's what I've got.

This is an ungeared 12V motor, and I'll have a go at wiring it up to thrash the carriage back and forth. We'll soon see what wears out first, and I'll be able to figure out some real-world speeds.

Vik :v)

This is an ungeared 12V motor, and I'll have a go at wiring it up to thrash the carriage back and forth. We'll soon see what wears out first, and I'll be able to figure out some real-world speeds.

Vik :v)

Indeed, DC motors do have a torque which is a function of the velocity. Ignoring friction, the torque is proportional to the current running through the motor; the current is proportional to the effective voltage applied; the effective voltage is the actual voltage minus the back-EMF (the voltage generated by the motor because it's also a generator).

Thus, when the motor velocity is such that the back-EMF equals the applied voltage, the motor produces no torque at all. If the motor velocity is zero, then all the applied voltage can be used for torque. And if the motor is running "backwards", you can get even more current and more torque (if you're not expecting this "extra" current, you can blow up your motor controller this way).

If you short the leads of a DC motor together, this is the same as applying a voltage of zero; then the torque will be non-zero if the motor is moving (and will be in the opposite direction of motion), so this acts as a brake.

Thus, when the motor velocity is such that the back-EMF equals the applied voltage, the motor produces no torque at all. If the motor velocity is zero, then all the applied voltage can be used for torque. And if the motor is running "backwards", you can get even more current and more torque (if you're not expecting this "extra" current, you can blow up your motor controller this way).

If you short the leads of a DC motor together, this is the same as applying a voltage of zero; then the torque will be non-zero if the motor is moving (and will be in the opposite direction of motion), so this acts as a brake.

Right, Ash and I have tried it for real. We used a nice, big 12V motor off an old HP printer of Adrian's, and put it on a makeshift carriage and bar assembly.

It moves things around at one heck of a speed. Far more than I need, and in desperate need of some PWM controlling!

We're going to need robust limit switches on this thing, probably backed up with an anchor and short chain.

Vik :v)

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It moves things around at one heck of a speed. Far more than I need, and in desperate need of some PWM controlling!

We're going to need robust limit switches on this thing, probably backed up with an anchor and short chain.

Vik :v)

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