How to Calculate the Intersection Point of Two Cycloids?

[physicsforumer]

Homework Statement

Greetings everyone!
I would like to know, how to calculate the intersection point of two cycloids.

Homework Equations

The equations of the cycloid are the following:
x=r(t-sin(t))
y=r(1-cos(t))

The Attempt at a Solution

I tried to solve it by myself, but I was'nt able to do it.
I did a research on the a topic, and I found that, the equations of the cycloids are transcendent, so it takes infinite steps to solve them.
But I can't imagine, that there is no solution for this problem. Can I get a little help with this? Thanks in advance!

 

[HallsofIvy]

There is no "general" way to find the intersections of all cycloids except to do the algebra. I don't know what you mean by "the equations of the cycloids are transcendent, so it takes infinite steps to solve them". Exponential and trig functions are transcendental functions and it is easy to solve them. However, we will need "special functions" to solve them just as we have to define the logarithm and inverse trig functions to solve such equations.

 

[physicsforumer]

Thanks for the immediate answer.
Yes, I can understand that, the intersection of a curtate or a prolate cycloid is very different.
As far as I know, transcendent in this case means that, there is no way to express t from the equation, that has t and sin(t) in it also. Because you would get in a loop of /sin(.) /arcsin(.) operations, while you wasn't able get closer to the solution.
For example:
t=sin(t) /arcsin(t)
arcsin(t)=t /sin(t)
t=sin(t) /arcsin(t)
...

Here is, how I tried to solve this:
To get the coordinates of IP(x,y) first I chose to deal with the x axis. I tried to make the equations of the two cycloids equal.
Cycloid #1: x=r(t-sin(t))
Cycloid #2: x=r(t-sin(t)+u)
If I try to make these two equations equal, I will get to the conclusion, that u=0, which is can't be true.
r(t-sin(t)))=r(t-sin(t)+u) /r
t-sin(t)=t-sin(t)+u /-t,+sin(t)
u=0
I'm afraid I'm missing something very basic here.

 

[physicsforumer]

Greetins again!

I checked, how parametric equations are solved.
Basically, you have to solve one of them for t.
y=r(1-cos(t))
t=arccos(-y/r+1)
I put this into the equation of x for both cycloid.
x=r(arccos(-y/r+1)-sin(arccos(-y/r+1)))
x=r(arccos(-y/r+1)-sin(arccos(-y/r+1))+u)
When I make these equations equal, I get to where I was before: u=0, which is can't be true in this case. Could you give me some advice, how to progress further with this?
Thanks in advance!

 

[sankalpmittal]

Greetins again!

I checked, how parametric equations are solved.
Basically, you have to solve one of them for t.
y=r(1-cos(t))
t=arccos(-y/r+1)
I put this into the equation of x for both cycloid.
x=r(arccos(-y/r+1)-sin(arccos(-y/r+1)))
x=r(arccos(-y/r+1)-sin(arccos(-y/r+1))+u)
When I make these equations equal, I get to where I was before: u=0, which is can't be true in this case. Could you give me some advice, how to progress further with this?
Thanks in advance!

Strange. You have cycloid equation in parametric form. That's one equation. All you can get is locus and talking about intersection point does not make any sense unless you have one more equation.

 

[BvU]

Dear pfr,

I have two things to remark:

1. what is the picture in post #1 representing ? A cycloid with the parametric description x=r(t-sin(t)), y=r(1-cos(t)) looks quite differernt in the x-y plane. What are XC and YC ?

2. You are starting to skid in post #4 when you think the second cycloid is described by
x=r(arccos(-y/r+1)-sin(arccos(-y/r+1))+u)
because that sets the t of cycloid #1 equal to that of cycloid #2 at the intersection point. There's no reason for that.

Picture shows blue line for x=t-sin(t) y=1-cos(t), red line for x+1.5=t'-sin(t') y=1-cos(t')
Clearly, when x1 = x2 then t is not equal to t'