20140711, 22:48  #1 
Sep 2006
The Netherlands
3^{6} Posts 
calculating with circles
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/wpcont...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 
20140711, 22:56  #2 
Sep 2006
The Netherlands
3^{6} Posts 
>> 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... 
20140711, 23:06  #3 
Sep 2006
The Netherlands
3^{6} Posts 
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! 
20140711, 23:22  #4 
"Mike"
Aug 2002
2^{3}×3×7^{3} Posts 
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/catalogsandresources/resources 
20140711, 23:31  #5  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
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 20140711 at 23:38 

20140711, 23:44  #6 
Sep 2006
The Netherlands
1011011001_{2} Posts 
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 :) 
20140711, 23:50  #7 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
I must admit though that I've never done a thing like CAD before.

20140712, 00:18  #8  
Sep 2006
The Netherlands
3^{6} Posts 
Quote:
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 youtcomes, which shouldn't happen :) So something still wrong in my math approach... 

20140712, 01:14  #9  
"William"
May 2003
New Haven
2×7×13^{2} Posts 
Quote:
(x  0.4)^2 + (y  0.195)^2 = 1 

20140712, 12:14  #10 
Sep 2006
The Netherlands
1331_{8} Posts 
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 20140712 at 12:29 
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