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 2014-07-11, 22:48 #1 diep     Sep 2006 The Netherlands 32×83 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/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
 2014-07-11, 22:56 #2 diep     Sep 2006 The Netherlands 32·83 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...
 2014-07-11, 23:06 #3 diep     Sep 2006 The Netherlands 32×83 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!
 2014-07-11, 23:22 #4 Xyzzy     "Mike" Aug 2002 2×4,139 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/catalogs-and-resources/resources
2014-07-11, 23:31   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by diep 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

 2014-07-11, 23:44 #6 diep     Sep 2006 The Netherlands 32×83 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 :)
 2014-07-11, 23:50 #7 science_man_88     "Forget I exist" Jul 2009 Dumbassville 838410 Posts I must admit though that I've never done a thing like CAD before.
2014-07-12, 00:18   #8
diep

Sep 2006
The Netherlands

32×83 Posts

Quote:
 Originally Posted by science_man_88 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...

2014-07-12, 01:14   #9
wblipp

"William"
May 2003
New Haven

32×263 Posts

Quote:
 Originally Posted by diep 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

 2014-07-12, 12:14 #10 diep     Sep 2006 The Netherlands 32×83 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 2014-07-12 at 12:29

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