is there any program to calculate 25 , 35 chains for frc

Iāve used this one for both belt and chain. For belt I just enter the number of millimeters instead of inches - it all scales.

http://www.botlanta.org/converters/dale-calc/sprocket.html

And OBTW, #25 is 0.25" pitch, #35 is 0.375" pitch. In general, US chain pitch is the tens place of the chain number times one eighth of an inch.

+1 to both the calculators linked above.

However, if you happen to want to make your own calculator (ie. for an excel sheet, in a cad document, etc.) the formula is as follows:

```
chain_center_distance = pitch * 1 / 8 * ( 2 * chain_links - sprocket_1_teeth - sprocket_2_teeth + sqrt( ( 2 * chain_links - sprocket_1_teeth - sprocket_2_teeth ) ^ 2 - ( 8 / ( ( PI ) ^ 2 ) ) * ( ( sprocket_1_teeth - sprocket_2_teeth ) ^ 2 ) ) )
```

Where CD is the center distance, p is pitch, l is the number of links in the chain, t1 is the number of teeth on one sprocket and t2 is the number of teeth on the other sprocket.

The unit you use for pitch will determine the output unit.

I have a C-C belt and chain calculator in my AMB Design Spreadsheet

Ā

Do you have a source or derivation for this formula? From what Iāve seen, the formula works out to be not numerically solvable for the C-C distance and needs to be solved numerically. See the derivation that I used in Section 4 and Appendix B of the equations paper for my spreadsheet

Derivation no, however it is the same calculation found in the source of the botlanta calc. I honestly have no idea how it works, just that itās what seems to be agreed upon by multiple popular online calculators.

The problem I see with your derivation is that chain doesnāt evenly wrap around the pitch circle of the sprocket. Instead of forming an arc, it forms many small chords between points on the pitch circle. Additionally, where the chain exits from either sprocket isnāt necessarily on the tangent line between the pitch circles, so your equation (93) isnāt necessarily correct.

I decided to attack the chain/belt calculation. First of all, my diagram:

Iām starting with lengths, as lower case variables, angles as greek letters. Link counts will be upper case letters, later on:

- l: overall length of belt/chain
- rā: pitch radius of larger pulley/sprocket
- rā: pitch radius of smaller pulley/sprocket
- c: center-to-center distance between pulleys/sprockets
- s: span of belt/chain on each side which does not contact pulley/sprocket
- Īø: angle [in radians] between āspanā and ācenter-to-centerā lines.

Clearly, l=2s+(\pi-2\theta)r_2+(\pi+2\theta)r_1=2s+\pi(r_1+r_2)+2(r_1-r_2)\theta

Further, the right triangle with hypotenuse c yields s^2=c^2-(r_1-r_2)^2 and sin\theta=(r_1-r_2)/c, allowing us to remove s and \theta.

Finally, l=2\sqrt{c^2-(r_1-r_2)^2}+\pi(r_1+r_2)+2(r_1-r_2)sin^{-1}(\frac{r_1-r_2}c)

To convert this to links, we note that l=L/p, where L is the link count of the chain and p is the pitch. Further, 2\pi r_1= T_1p and 2\pi r_2= T_2p, where T_1 and T_2 are the number of teeth on the larger and smaller sprockets/pulleys respectively. Dividing both sides by p then gives:

L=2\sqrt{(\frac{c}{p})^2-\frac{(T_1-T_2)^2}{4\pi^2}}+\frac{T_1+T_2}{2}+\frac{(T_1-T_2)}{\pi}sin^{-1}(\frac{p(T_1-T_2)}{2\pi c})

This is solvable for c, but it clearly does not produce the same answer @a_cool_username presented.

Yesā¦ this approach produces a L that is too long or a c that is too short, with errors growing as the tooth count approaches zero.

Unfortunately the geometry is not so curvaceous as you and Ari have drawn. Consider this arrangement of two 4-tooth sprockets with a 10-link loop of chain between them. I have superimposed two states as the system rotates 45 degrees. Notice how the effective sprocket radius varies between the two states.