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Old 2014-07-11, 22:48   #1
diep
 
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hi!

I'm trying to draw in cad in a 'mathematical correct" manner a small part for 3d printer for GT2 belt.

Yet i probably missed some lessons in math here and there.

A good example of the picture is here: http://www.adafruit.com/blog/wp-cont...2/gt2tooth.jpg

I'm trying to get the math right for 1 part of a tooth.

As we see we have 3 circles, yet i just worry about the 2 larger ones.
Let's focus upon the circle which i'll put at O :
That one had diameter r.

y ^ 2 + x ^ 2 = 0.555 ^ 2

then if i interpret the drawing correct the other circle is at

(x + 0.4) ^ 2 + (y - 0.195) ^ 2 = 1

Of course for the same tooth there is also y + 0.195

If i try to calculate now where the circles touch each other, then i get in troubles with limited mathskills...

Where is the mistake?

I'm first trying to write it in the highschool form of y = ...

I get then to : y = +- sqrt( -x^2 + 0.555^2 )
and y = +- sqrt( -x^2 -0.8x + 0.84) + 0.195

Now i want to solve this to see where they hit.
I square both y's then and get to:

-x^2 + 0.555^2 = - x^2 - 0.8x +0.84 + 0.195^2 +- 0.195 sqrt(-x^2 -0.8x + 0.84)
<=> 0.8x -0.57 = 0.195 sqrt( -x^2 -0.8x + 0.84)
now squaring that again i get :
(0.8x -0.57)^2 = 0.195^2 * ( -x^2 -0.8x + 0.84)

0.678025 x ^ 2 - 0.88158x + 0.292959 = 0

D = b^2 - 4 ac = -0.17350808

Now my skills working with i are not so high as i get then to:

x1,x2 =( 0.88158 +- 0.131722464 i ) / (2 * 0.678025)

In short i didn't manage to find where the 2 circles touch each other (and where they should also have the same vector going up)

Anyone?

Thanks,
Vincent
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Old 2014-07-11, 22:56   #2
diep
 
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>> -x^2 + 0.555^2 = - x^2 - 0.8x +0.84 + 0.195^2 +- 0.195 sqrt(-x^2 -0.8x + 0.84)

checking now whether that would be

-x^2 + 0.555^2 = - x^2 - 0.8x +0.84 + 0.195^2 +- 2 * 0.195 sqrt(-x^2 -0.8x + 0.84)

and what the result is then...
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Old 2014-07-11, 23:06   #3
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Interesting!

Now the rest of sequence i get is :

0.7921 x ^ 2 - 0.79032 x + 0.197136 = 0

D = b^2 - 4 ac = 0

x = 0.79032 / (2 * 0.7921 ) = 0.498876404

Maybe highschool math saved me again!
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Old 2014-07-11, 23:22   #4
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No idea if this helps, but Gates has a pile of technical documentation for belts. (We got lost there one weekend!)

http://www.gates.com/catalogs-and-resources/resources
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Old 2014-07-11, 23:31   #5
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Quote:
Originally Posted by diep View Post
As we see we have 3 circles, yet i just worry about the 2 larger ones.
Let's focus upon the circle which i'll put at O :
That one had diameter r.
really I see four minimum from what I looked at, the following is my reasoning:

4*0.15 +0.555 +0.555 = 0.6+1.11 = 1.71< 2 accounting for four circles and edit:"in need" a gap between two of them.

edit2: and the middle part in the drawing looks flat.

Last fiddled with by science_man_88 on 2014-07-11 at 23:38
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Old 2014-07-11, 23:44   #6
diep
 
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Taken a look, thanks for the link!

Interesting for later considerations if i build a CNC myself (which i intend to) with respect to tolerances and forces, they have for example something about GT3 as it seems (similar to GT2).

No document theirs i found so far seems to have any of the math required to design something like that in CAD :)
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Old 2014-07-11, 23:50   #7
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I must admit though that I've never done a thing like CAD before.
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Old 2014-07-12, 00:18   #8
diep
 
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Quote:
Originally Posted by science_man_88 View Post
really I see four minimum from what I looked at, the following is my reasoning:

4*0.15 +0.555 +0.555 = 0.6+1.11 = 1.71< 2 accounting for four circles and edit:"in need" a gap between two of them.

edit2: and the middle part in the drawing looks flat.
Yeah the belt has flat parts in between the teeth indeed.

I'm taking a look at just 1 slope of one of the teeth, as logically it's mirrored at the other side of the tooth.

and the teeth end consists out of a circle with diameter 0.55 mm and part of the slope is a small segment from a circle with diameter 1.00 mm

My math is about that, and the outcome 0.498876404 if i fill that in, in both formula's i have, i seem to get different y-outcomes, which shouldn't happen :)

So something still wrong in my math approach...
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Old 2014-07-12, 01:14   #9
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Quote:
Originally Posted by diep View Post
As we see we have 3 circles, yet i just worry about the 2 larger ones.
Let's focus upon the circle which i'll put at O :
That one had diameter r.

y ^ 2 + x ^ 2 = 0.555 ^ 2

then if i interpret the drawing correct the other circle is at

(x + 0.4) ^ 2 + (y - 0.195) ^ 2 = 1
If you are using the usual convention that up is positive y and right an positive x, the second circle's center is at (.40, .195). It's equation is

(x - 0.4)^2 + (y - 0.195)^2 = 1
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Old 2014-07-12, 12:14   #10
diep
 
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hi wblipp,
that is indeed one of the 2 slopes.

As for the math it seems that there is another manner to calculate it.

If we take the length from the r=1.00 circle to the r=0.555 circle,
and we already know that its position is (x,y) = (-0.4,0.195)

Then the distance of that i noticed is sqrt( 0.195^2 + 0.4^2 ) = 0.445

Which happens to be 1 - 0.555

So the line that defines the point where the circles touch goes from origin of the 1.00 circle through the origin of the 0.555 circle

In short we can then easily calculate its position as distance from origin of the 0.555 circle:

(555/445) * {x = 0.4 ,y = -0.195} = {0.498876404 , -0.243202247}

If i fill in then into: x^2 + y^2 = 0.555^2 ==> y^2 = 0.555^2 - 0.498876404^2
==> y = +- 0.243202247

Now let's fill in in the formula :

(x +- 0.4) ^2 + (y - 0.195) ^2 = 1 <==> y^2 - 0.39y + 0.195^2 - 1 + (x +- 0.4) ^2 = 0

Then let's calculate for both +0.4 as well as -0.4 ==>
(edit: with better numbers...)
A : y^2 - 0.39y + 0.195^2 - 1 + 0.807978791 = y^2 - 0.39y - 0.153996209..
B : y^2 - 0.39y + 0.195^2 - 1 + 0.009776543 = y^2 - 0.39y - 0.952198457..

D = b^2 + 4ac ==>
A: D = 0.768084838..
B: D = 3.960893827..

A: x1,x2 = (0.39 +- sqrt(D)) / 2 = (0.39 +- 0.876404494.. )/2 = { -0.243202247 , 0.633202247 } one solution outside the 0.555 circle
B: x1,x2 = (0.39 +- sqrt(D)) / 2 = (0.39 +- 1.990199444...)/2 = { -1.600199444 / 2 , 2.380199444 / 2} both solutions outside the 0.555 circle

So that's the answer i was looking for (after the edit).

Last fiddled with by diep on 2014-07-12 at 12:29
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