# Telescope mirror polishing simulator

1. Mar 6, 2004

### Lerch

Greetings All,

I am attempting to develop a software simulation of the dynamics involved in machine figuring concave parabolic telescope mirrors. As a member of the amateur telescope making community, we currently don't have any tools to do this other than the "trial and error" method.

You may be asking "So, what brings you here?" Well, long story short, I'm trying to model the situation, and I've already ran into problems and I need some guidance. Before I can ask my question, I need to describe the situation.

I have a polishing machine that looks like this:

http://lerch.no-ip.com/atm/Projects/Polisher/page_01.htm

The big piece of glass on the bottom is the telescope mirror which is 317mm in diameter. The mirror is on a turntable that spins at a fixed angular velocity. On top of the mirror is a polishing tool, 196mm in diameter. The tool is free to rotate about its center.

For the sake of keeping the initial simulation as simple as possible, I've assumed the following conditions:

#1 The surface of the mirror and tool are flat instead of curved
#2 The center of the tool is held at a fixed radius from the center of the mirror.
#3 The coefficient of kinetic friction is constant for all velocities.
#5 The angular velocity of the mirror is -4.189 radians per second
#6 At startup the tool has no angular velocity

My first goal for the simulation, "Predict the angular velocity of the tool" (which seemed simple enough).

First, I created a table elements, each containing the force tangent to the center of the tool:
http://lerch.no-ip.com/atm/Tang_Force.jpg

Next, I multiplied each table element by the radius in mm of the tool:

Then I found the sum of the table, multiplied that by the force of friction, and divided the product by the inertia of the tool to give an initial angular acceleration. Thus completes time = 0, and now I'm stuck....

At time = 1, I have a tool angular velocity equal to the tool angular acceleration at time = 0, yes?

I know my force table calculations have to change, other wise I'd get infinite acceleration. I know that the change is related to the angular velocity for each point on the tool, compared to the angular velocity of the mirror at the same point., and this is where I'm stuck. I don't know how to make this comparison.

Any Suggestions on how to proceed?

Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)

"Anything that can happen, will happen" -Stephen Pollock from:
"Particle Physics for Non-Physicists: A Tour of the Microcosmos"

" Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent. "
Calvin Coolidge

2. Mar 6, 2004

### NateTG

Yes, with the assumptions that you're making, you'll get constant acceleration. There are essentially three states - constant acceleration in two directions, and a steady state at some rpm where the net torque of friction is zero.

Depending on your model for friction, you should be predicting constant angular acceleration.

The normal equations are:
$$\theta=\theta_0+\omega_0 t + \frac{1}{2}\alpha t^2$$
$$\omega=\omega_0 + \alpha t$$
$$\theta=\theta_0+\frac{1}{2}(\omega+\omega_0)t$$

As the model for mirror shapes and friction becomes more sophisticated, the problem rapidly approaches the numerical realm where you can use the same equations:
$$\theta(t + \epsilon)=\theta(t)+\omega(t) \epsilon + \frac{1}{2}\alpha(t)^2$$
and
$$\omega(t+\epsilon)=\omega(t)+\alpha(t)\epsilon$$

This should all seem quite familiar if you've taken an introductory mechanics course.

Last edited: Mar 6, 2004
3. Mar 6, 2004

### Lerch

Hello NateTG,

While I'm doing "ok" with the angular acceleration and velocity problems, I don't know how to solve for the torque on each element representing tool (when both the tool and mirror are rotating.)

I'm pretty sure I've got the method for solving torque on each element representing the tool when the tool is NOT rotating, but I can't grasp the method needed for when the tool IS rotating..

I keep track of the torques on the tool in an array. Each element of the array represents 1 mm^2, with the origin of the array at the center of the tool.

BTW, I am in way over my head on this, but learning quickly. I lack a formal education in physics or mechanics.

Thanks,
James

4. Mar 6, 2004

### NateTG

$$\vec{\tau}=\vec{f} \times \vec{r}$$

where r is the lever arm, f is the force, and x is the cross product.

Torque is equal to the length of the of the lever arm multiplied by the magnitude of the force multiplied by the sine of the angle measured from the force to the lever arm. (Since sine is an odd function this is important.)

You can probably start out by assuming that the force of friction is acting tangentially to circles of revolution on the large disk.

5. Mar 6, 2004

### Lerch

Ok, here's the bit that I get lost on: (step D,below)

For every point (X,Y) on the surface of the tool (small disk) in contact with the mirror (large disk). The torque at any point (X,Y) on the tool is the product of:

A) The radius in mm for this point(X,Y) of the tool
B) sin (perpendicular angle between mirror center, this point, and tool center)
C) Force of kinetic friction
D) ??? (something that off-sets this torque when mirror reaches a steady state angular velocity, neither accelerating or decelerating)

I'm currently thinking "D" is a change in sign for torque at this point. I'm supposing that when the angular velocity of the tool at this point is greater than the angular velocity of the mirror underneath, we need to change the sign of the torque at this point...

If my supposition is correct, I don't know how to perform the comparison.

If it would help, I could post some code snippets showing either:
A) what I'm trying to do
B) My Ignorance :)

BTW, thank you for your time, I really do appreciate it!

James

6. Mar 7, 2004

### NateTG

I should really figure out how to post drawings.

It seems like it might be usefull to keep track of locations in terms of the center of the mirror, and the center of the tool.

The first thing that would be nice to figure out is translation between the two systems.

Since we can, we might as well set up the coordinates so that the center of the mirror is $$(0,0)$$.

Let's throw out some constants:
$$d$$ the (planar) center to center distance from the tool to the mirror.
$$\vec{d}$$ is the vector between the center of the mirror, and the center of the tool.
$$\vec{r}_{mirror}, \vec{\omega}_{mirror}$$ the radius and angular velocity of the mirror.
$$\vec{r}_{tool},\vec{\omega}_{tool},\vec{\alpha}_{tool},I_{tool}$$ are the angular speed, velocity and acceleration, and the moment of inertia of the tool of the tool.

Now, the local velocity of the mirror is
$$\vec{v}_{mirror}(\vec{r})=\vec{\omega}_{mirror} \times \vec{r}$$

The local velocity of the tool is
$$\vec{v}_{tool}(\vec{r})=\vec{\omega}_{tool} \times (\vec{r}-\vec{d})$$

The local force of friction on the mirror is
$$\vec{f}(\vec{r})=\frac{\vec{v}_{mirror}-\vec{v}_{tool}}{|\vec{v}_{mirror}-\vec{v}_{tool}|} \mu_k f_{normal}$$
where $$\mu_k$$ is the coeeficient of kinetic friction, and $$f_{normal}$$ is the normal force on the bit of area that you're working with. Since that part is a (fiddle friendly) scalar, it's not such a huge deal.

Finally:
$$\tau_{net}=\int \vec{f}(\vec{r}) \times (\vec{r}-\vec{d}) d\vec{r}$$
$$\alpha_{tool}=\frac{\tau_{net}}{I_{tool}}$$

It may make more sense to center the system on the middle of the tool. Let me know if that helped any.

7. Mar 8, 2004

### Lerch

Well, your post helps, but I run into two educational problems

A) I'm working on replying in Latex images (forgive errors as I learn)

B) I don't 'get' vectors yet... I kind of get vectors, but not entirely,

For instance If I see

$$\vec{a}-\vec{b}=\vec{c}$$
and

$$\vec{a}=(1,2)$$
$$\vec{b}=(3,4)$$

then $$\vec{c}=(1-3,2-4)$$
or $$\vec{c}=(-2,-2)$$

Also, if I see:
$$|\vec{c}|$$

then were looking for the square root of the sum of the squares. (aka the magnitude of the vector?)
$$\vec{c}=(-2,-2)$$
then
$$|\vec{c}|=\sqrt{-2^2+-2^2$$

(assuming I'm on track, and not totaly ignorant of vectors, here's where I get confused:

$$\vec{v}_{mirror}(\vec{r})=\vec{\omega}_{mirror} \times \vec{r}$$

Given:
$$\vec{\omega}_{mirror}=-4.1$$

$$\vec{r}=(50,50)$$

does this mean: (?)
$$\vec{v}_{mirror}(\vec{r})=(-205,-205)$$

I'll stop here, for if I'm wrong then there is no point in proceeding untill I continue my education of vectors :)

Thanks,
James

8. Mar 8, 2004

### NateTG

You've got it so far.

The curve ball that I threw is that angular velocity, and acceleration are vectors that are parralel to the axis of rotation. (Yeah I know it's a bit wierd.) This is usefull because it allows them to be used in cross products.

So, the cross product of two 3 dimensional vectors is:

$$(a_i,a_j,a_k) \times (b_i,b_j,b_k)=(a_jb_k-a_kb_j, a_kb_i-a_ib_k,a_ib_j-a_jb_i)$$

$$\vec{a} \times \vec{b}$$ is always perpedicular to both $$\vec{a}$$ and $$\vec{b}$$

Now, in the example:
$$\vec{v}_{mirror}(\vec{r}) = \omega_{mirror} \times \vec{r}$$
$$\vec{v}_{mirror}( <50,50,0> )= <0,0,4.1> \times <50,50,0> =$$
$$<0 * 0 - 4.1 * 50,4.1*50 - 0* 0, 0*50-0*50>=<-205,205,0>$$
So the velocity of the surface of the mirror at $$(50,50)$$ is $$(-205,205)$$
which makes sense since postive angular velocities are considered to be anti-clockwise.

Does that help at all?

Last edited: Mar 8, 2004
9. Mar 8, 2004

### Lerch

I'll need to chew on this info for a bit, vectors are still a new concept to me.

Back in a bit, hopefully with more progress and questions :)

James

10. Mar 12, 2004

### Lerch

Its me again :)

I actually made great progress the last few days, thank you for your help! I actually have something that works, but now I have a new problem :0

The simulation runs true with measured results UNTILL parts of the tool hang off the mirror. The simulator says to the tool will slow down, in reality the tool speeds up.

After thinking about it for a while, I realized my error. Current code assumes the normal force / unit area is constant across the tool while working out the local force of friction. This is true while the entire tool is on the mirror, this is FALSE whent he tool overhangs the mirror. We now have a weight hanging off the mirror trying to pivot around the edge of the mirror. Now I have to re-calculate the normal force for each point across the whole tool, if any of the tool overhangs the mirror :0

You wouldn't have a nifty formula handy for calculating weight distribution across a disk, if part of the disk is overhanging a larger disk?

"Oh Boy" :)

James

11. Mar 12, 2004

### NateTG

Ugh, I'm not sure if I have anything *nice* that will handle all of the cases, but I'll think about it. You should BTW also check to make sure that you're not calculating torque on the open area.

I'll try to get back to you later tonight.

12. Mar 12, 2004

### NateTG

For now what you can do is to add up the torque first, then multiply the total by the normal force and the force of friction instead of doing it bit by bit. Then you can handle the 'overhang' problem by simply checking the radii when you do the net torque calculation numerically.

Edit:
This isn't exactly clear. What you can do is numerically determine the total area that is in contact so you can get the pressure due to the normal force.

Since you're already doing a numeric operation on the tool surface, it shouldn't be a whole more work.

Last edited: Mar 13, 2004
13. Mar 16, 2004

### Lerch

Hi Nate,

Perhaps a simplification of my current problem is in order. Below is a link to an image showing two situations for the tool(brown) and mirror(blue)

In the top situation, the normal force for friction between the two items should be uniform for each point measured.

In the bottom situation, common sense says the normal force should be greater for points toward the right side of the contact area, and lower for points toward the left side of the contact area.

The problem reminds me of the physics example of "two window washers standing on a platform supported by two cables, wondering how much weight each cable is supporting as they walk about platform"

Thanks,
James

14. Oct 21, 2006

### wingfloat

go figure!

Wow!

I just found this sight. It seems that a simple thing is being complicated to the extreme i.e. figuring mirrors , like every thing else these days .

Foucault test, plot the zonal errors, then proportionally transfer to the polishing tool .