Determine the diameter of a circle with constraints

Question

I would like to solve in analytical way the following problem:

1. I have a circle with diameter D1=70mm
2. I have another circle with diameter D2=Xmm which is my unknown of this problem
3. The center to center distance between the circles is 250mm
4. The length of the "belt" (depicted in blue line color) is always fixed to 800mm

The circle identified by diameter D1 represents the driving pulley. The circle identified by diameter D2 represents the driven pulley. The blu line is always tangent to both circles (pulleys) and represents the belt.

Now, when I change the diameter D1, I will obtain another value of diameter D2 and today I have to determine the solution with the help of a parametric CAD drawing.

Is there a way to solve it analytically (for example using an Excell sheet).

Thank you in advance!

Deatil of the winding angle on the driven circle:

winding angle

Edit: the winding angle is the same in both circles: detail of the system

Answer 1

\begin{align} |T_tT_2|=|O_1C| &= \sqrt{d^2-(r_2-r_1)^2} \tag{1}\label{1} ,\\ \phi&=\arcsin\Big(\frac{r_2-r_1}d\Big) \tag{2}\label{2} . \end{align}

Constraint:

\begin{align} (\tfrac\pi2-\phi)\cdot r_1 +\sqrt{d^2-(r_2-r_1)^2} +(\tfrac\pi2+\phi)\cdot r_2 &=\tfrac12\,L \tag{3}\label{3} , \end{align}

\begin{align} \text{or }\qquad \left( 2\,(r_2-r_1)\arcsin\left(\frac{r_2-r_1}d\right) - (L-\pi\,(r_2+r_1) -2\,\sqrt{d^2-(r_2-r_1)^2}) \right)^2 &=0 \tag{4}\label{4} . \end{align}

Unfortunately, in general, given $r_1,d$ and $L$ there is no analytic solution for $r_2$ in \eqref{4}, so we need to use numeric methods.

For example, in Excel, you can type the formula \eqref{4} and use built-in non-linear solver to minimize it.

For example, for $r_1=35,\ d=250,\ L=800$ we get $r_2=59.71462$.

As the first approximation to $r_2$ you can try

\begin{align} r_{2\,(0)} &= r_1+\pi\,d-\frac{\pi^3\,d^2}{d\,(\pi^2-2)+L-2\pi\,r_1} . \end{align}

For the above example, $r_{2\,(0)}\approx 59.69$, which is pretty close.

For the numeric approximation you can use, for example, Halley's method as a root-finding algorithm:

\begin{align} r_{2(n+1)} &= r_{2(n)} - \frac{2\,f(r_{2(n)},r_1,d,L)\,f'(r_{2(n)},r_1,d)} {2\,f'(r_{2(n)},r_1,d)^2-f(r_{2(n)},r_1,d,L)\,f''(r_{2(n)},r_1,d)} , \end{align}

where

\begin{align} f(r_{2},r_1,d,L) &= 2\,(r_2-r_1)\,\arcsin\Big(\frac{r_2-r_1}d\Big) -L+\pi\,(r_2+r_1)+2\,\sqrt{d^2-(r_2-r_1)^2} ,\\ f'(r_{2},r_1,d) &= 2\,\arcsin\Big(\frac{r_2-r_1}d\Big)+\pi ,\\ f''(r_{2},r_1,d) &= \frac 2{\sqrt{d^2-(r_2-r_1)^2}} . \end{align}

$$ \begin{array}{cc} \hline n & r_{2(n)} \\ \hline 0 & 35.0000000000 \\ 1 & 59.6915126334 \\ 2 & 59.7146200516 \\ 3 & 59.7146200510 \\ \hline \end{array} $$

This is a minimal python example:

from math import *
def f(r2,r1,d,L) : 
  """
  r2 - radius of the driven pulley
  r1 - radius of the driving pulley
   d - the center to center distance between the circles 
   L - length of the belt
  """
  return 2*(r2-r1)*asin((r2-r1)/d)-L+pi*(r2+r1)+2*sqrt(d*d-(r2-r1)**2)
def df(r2,r1,d) : 
  """
  f'(r2)
  """
  return 2*asin((r2-r1)/d)+pi
def ddf(r2,r1,d) : 
  """
  f''(r2)
  """
  return 2/sqrt(d*d-(r2-r1)**2)
def F(r2,r1,d,L) :
  """
  next approximation to r2
  """
  vf=f(r2,r1,d,L)
  vdf=df(r2,r1,d) 
  vddf=ddf(r2,r1,d) 
  return r2-2*vf*vdf/(2*vdf**2-vf*vddf)
def calc_r2(r1,d,L,eps=1e-6) :
  r2o=r1
  r2=F(r2o,r1,d,L)
  while(abs(r2-r2o)>eps) :
    r2o=r2
    r2=F(r2o,r1,d,L)
  return r2

print(calc_r2(35,250,800))
# 59.71462005113761
print(calc_r2(20,250,800))
# 72.03312874960544
print(calc_r2(3,250,800))
# 84.05267790667017

Answer 2

Hint - Suppose we consider the case that D2 > D1

Now, drop the perpendicular from centre of D1 to the radius connecting tangent point and centre of D2. You will get a right triangle with arm lengths 250, $\frac{x-70}{2}$

Can you relate the angles of this triangle to the angle subtended by the sectors made by rope contact with each circle? And then use the total length equation to get an equation in x?

EDIT

The entire diagram is symmetric about the line joining the centres

Let $C_1$ be centre of circle 1, $C_2$ be centre of circle 2, and the foot of perpendicular be P. Hence, angle subtended by contact line on small circle is $2*(90 - \angle PC_1C_2)$

For $C_2$, the angle subtended is $2*(180-\angle PC_2C_1)$

To see this, just extend the line joining the two centres till it intersects the circle. Once you do that, it should not be difficult to relate the angles as above