So with pullies at the bottom corners, the motors would provide some tension to help pull sag out of the long chain and offset friction? That could help even if the cable wasn’t being used to measure as well.

# Draw-Wire Calibration/Positioning System

**madgrizzle**#82

I think the tension in the segment of wire from the top pulley to the fixed point on the sled and the segment of wire from the bottom pulley to the spool/motor/encoder would be, for lack of better word, similar. I don’t want to say equal without having math behind it, but in a static condition the horizontal component of the * combined* tension of the left two cables would equal the horizontal component of the

**combined**tension of the right two cables (not shown in the drawing). The vertical component of the

**combined**tension of the top cables would equal the vertical component of

**combined**tension of the bottom cables plus effect of gravity. Yes?

Either this arrangement will pull itself into the bottom corner and improve upon the way its done today, or the tensions between the top cable and bottom cable somehow/basically cancel each other out in such a way that it results in no improvement. I haven’t figured out the answer, so that’s why I ask.

**dlang**#83

so how would this allow the angles to change on the two sides? as you move up

and down, the angles of the lines are going to change and the two lengths of

line (top of sled to top of pully and bottom of pully to motor) are going to end

up being different lengths.

Please draw this out in the basic 9 positions (top middle bottom x left middle

right) and see what the line lengths and angles would be.

David Lang

**madgrizzle**#84

For the first case, would not the length from pulley to the center of the router bit equal the length from pulley to top spot + pulley to bottom spot / 2 ? (Assuming for sake of discussion the pulley diameter was zero). And those two lengths added together is known.

Not sure about the second case, but I think it’s calculatable as well… will have to ponder it.

**dlang**#85

No, think of a simple triangle with all angles 60 degrees and all sides with a

length of 2.

the distance from the top of the triangle to the middle of the bottom is

sqrt(2).

the two lengths are not known, the sum of them may be known (to the extent that

we know the length of the chains today, sag and stretch come into play here as

well)

David Lang

**madgrizzle**#86

Yep good point (but should be sqrt(3) I think). So the question becomes if you can determine the distance if all you know is the sum of the lengths of two sides and the length of the remaining one… doesn’t sound like it unless you have an angle as well or the sled doesn’t rotate.

**madgrizzle**#87

I wonder if the sled won’t rotate with how cables are connected… If that’s the case, I *think* you have enough info to so the trignometry.

**dlang**#88

ignoring for the moment the diagram with a second line to the bottom of the

machine (I don’t fully understand that one), what is there to keep it from

rotating?

the pully up in the top corners sure won’t.

David Lang

**blurfl**#89

The bricks and the same rig in the opposite direction should be sufficient to stabilize it.

**blurfl**#90

If you’re measuring the total amount of cable between the motor on the sled and the fixed end, won’t that length be the same for a given sled location regardless of the sled rotation?

**dlang**#91

I don’t think so. as the triangle gets narrower, it gets taller.

the extreme cases

sled mounts square to the pully, sides and sled all 2 (perimeter of the triangle

is 6) , distance from pully sqrt(3)

all three in line, total perimeter is still 6, but the distance from the pully

to the center of the sled is 4

David Lang

**madgrizzle**#92

Yes, but if the sled rotates you need one more piece of info to calculate the length from the pulley to the center of router bit. Maybe it’s calculatable in a sense there may only be one solution when both sides are hooked up… just theorizing

**dlang**#93

Tension in each direction varies drastically, so the rig in the opposite

direction will not help much.

bricks help some, but only if the tension is really balanced to start with.

since the pully is not friction free, as the motor starts pulling, it will tend

to pull the bottom more than the top.

But this is why I suggested drawing this out and checking dimensions in a 9

point sample grid.

David Lang

**dlang**#95

Could be. I’m not seeing it, but I’m not sure.

That’s why I asked for the calculation based on the 9 points.

I’ve been wrong about a lot of things (some that I thought were good ideas that

didn’t work, some that I thought were bad ideas, but worked), so I voice that

I’m skeptical, but try to identify ways to test to be sure.

I think in this cast, top center (both sides the same), rotate the sled and it

will drop as the effective length increases.

David Lang

**madgrizzle**#96

Since I can’t use computer and have nothing better to do, here’s another theory. If the sled can rotate and not move the center of rotation, then you can assume, **for the purpose of calculating the distance from the pulley to the router bit**, that the two sides are the same length. Therefore, you know the hypotenuse of a right angle and the length of one other side… so you have to the info you need.

**blurfl**#97

Bingo. Move the attachment points on the sled out to the circumference of the respective pulley and make all three the same radius, and all three pullies drop out of the calculation.

**madgrizzle**#98

I think this is generic no matter where the sled is located if the assumption is correct that you can assume, for the purpose of calculating the distance between the top pulley and the center of the router bit, that the sides of the triangle are equal.

A is known because it’s the distance between the top fixed pulley and the target coordinates.

B is known because it’s the fixed spacing between the two sled points

Angle “a” is known because it equals atan(B/2/A)

F is known because it’s the difference in the Y coordinates of the top pulley and the target.

Angle “c” is known it equals acos(F/A)

Therefore, all others become calculatable and the total length of cable to be fed out is equal to L = E + A + D

**madgrizzle**#99

And this math assumes the pulleys have a zero diameter. Additional math is needed to work through what’s obviously a bad assumption.

**madgrizzle**#100

I apologize for multiple posts, but those on email don’t see edits…

I think my statement that “for the purpose of calculating the distance from the pulley to the router bit” is wrong, because that distance is known and calculable… what I should have said was “for the purpose of calculating the total length of wire/cable/chain”

To explain the theory about assuming that both sides of the triangle are equal, the idea is that if the sled can rotate freely, then the total length of wire/cable/chain cannot change. If it had to change, it could not rotate freely. Therefore, L = L no matter if the sled rotation is 0 or if it’s 90 degrees and whatever L equals when both sides of the triangle are equal is the same L if both sides of the triangle are not equal. So assume its equal in order to determine L in a simpler manner.