# Is it possible to solve for “t?”

• I
Devin-M
TL;DR Summary
This formula has to do with finding the quickest route between 2 horizontal points—

A=-(B/2(pi-t+sin(t)))(1-cos(t))

^When B and t (theta) are known we can find A, but when A and B are known, is it possible to find t (radians)?
This formula has to do with finding the quickest route between 2 horizontal points—

A=-(B/2(pi-t+sin(t)))(1-cos(t))

^When B and t (theta) are known we can find A, but when A and B are known, is it possible to find t (radians)? I’ve been told it isn’t. Is that true? Why or why not?

Mayhem
Hello. Is this the formula? $$A = -\frac{B}{2} (\pi-t+\sin{(t)})(1-\cos{(t)})$$ Sorry, it's just a little hard to be sure without proper syntax.

• Devin-M
Mentor
This formula has to do with finding the quickest route between 2 horizontal points—

A=-(B/2(pi-t+sin(t)))(1-cos(t))

^When B and t (theta) are known we can find A, but when A and B are known, is it possible to find t (radians)? I’ve been told it isn’t. Is that true? Why or why not?
I don't believe it's possible to find an exact solution for t by algebraic means (notwithstanding the Lambert w-function).

However, it would be possible to find an approximate solution numerically, using any of several techniques, such as Newton-Raphson or others.

BTW, I'm not sure what your equation is, as already noted by @Mayhem.

• Devin-M
Devin-M
Hello. Is this the formula? $$A = -\frac{B}{2} (\pi-t+\sin{(t)})(1-\cos{(t)})$$ Sorry, it's just a little hard to be sure without proper syntax.

I added an extra set of parenthesis:

##A = -(B/(2((pi-t)+sin(t))))(1-cos(t))##

When B and t are 1, A should be -0.077...

I added an extra set of parenthesis:

##A = -(B/(2((pi-t)+sin(t))))(1-cos(t))##

When B and t are 1, A should be -0.077...

As others have said, it can be solved numerically. However, it has multiple solutions. For A=-0.077 and B=1.0, I find 6 solutions for t, one of them being t=0.999665. How do you know which one you want?

Devin-M
The formula is intended to state, for example, “when an inverted, truncated cycloid has an initial generating angle of 1 radian (t) and final angle of 2pi-t, and the horizontal distance between the truncated cusps is 1 (B), then the depth at the truncated cusps is -0.077... in the Y axis.

Mayhem
My intuition tells me that it might have some solutions for ##t \in \mathbb{C}## since that allows us to rewrite trigonometric functions in all but boring ways, but I doubt that would have much practical application as complex radians isn't a subject I'm very familar with!

• Devin-M
Mentor
My intuition tells me that it might have some solutions for ##t \in \mathbb{C}## since that allows us to rewrite trigonometric functions in all but boring ways, but I doubt that would have much practical application as complex radians isn't a subject I'm very familar with!
With t appearing both inside and outside trigonometric functions you won't find a nice solution (excluding a few special cases). You can rewrite them as exponential functions but then you still have t and exp(t) which is the same problem.
##A = -(B/(2((pi-t)+sin(t))))(1-cos(t))##
Re-formatted:
##A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)} ##

• Devin-M
Mayhem
With t appearing both inside and outside trigonometric functions you won't find a nice solution (excluding a few special cases). You can rewrite them as exponential functions but then you still have t and exp(t) which is the same problem.Re-formatted:
##A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)} ##
Just a guess - could you maybe rewrite it and use the Lambert-W function?

• Devin-M
Devin-M
The other helpful part of the solution is:

##r=B/(2((pi-t)+sin(t)))##

Where r is the cycloid generating radius, B is the horizontal distance between the truncated cusps, and t is the initial cycloid generating angle in radians.

So if B and t are both 1, it’s equivalent to saying “when the initial inverted cycloid generating angle is 1 radian (t), the final angle is 2pi-t, and the horizontal distance between the truncated cusps is 1 (B), then the cycloid generating circle radius is 0.167...” and from the other equation “the depth at the truncated cusps is -0.077... (A) in the Y axis.”

Homework Helper
Gold Member
The other helpful part of the solution is:

##r=B/(2((pi-t)+sin(t)))##

Where r is the cycloid generating radius, B is the horizontal distance between the truncated cusps, and t is the initial cycloid generating angle in radians.

So if B and t are both 1, it’s equivalent to saying “when the initial inverted cycloid generating angle is 1 radian (t), the final angle is 2pi-t, and the horizontal distance between the truncated cusps is 1 (B), then the cycloid generating circle radius is 0.167...” and from the other equation “the depth at the truncated cusps is -0.077... (A) in the Y axis.”
Is this a second simultaneous equation that must be satisfied? Is ##r## known? If so, this looks easier to solve. If not, how is it helpful?

Devin-M
Is this a second simultaneous equation that must be satisfied? Is r known? If so, this looks easier to solve. If not, how is it helpful?

From B and t, one can calculate A and r, but from A and B it becomes very difficult or impossible to find t and r exactly, though they can be found to any arbitrary desired precision, I suspect?

Homework Helper
Gold Member
From B and t, one can calculate A and r, but from A and B it becomes very difficult or impossible to find t and r exactly, though they can be found to any desired precision, I suspect?
So the new equation for ##r## does not help to solve the first equation for ##t##. It is just another result, once ##t## is known.

Devin-M
Correct.

Devin-M
It’s ironic because the implication is that calculating the “quickest” route between 2 horizontal points takes an infinite amount of time.

The “quickest” path between 2 horizontal points has infinite complexity, etc...

Devin-M
Just to confirm, would it be accurate to make the following statement—

“Whenever t & B are rational, A is irrational. Whenever A & B are rational, t is irrational.”

?

Mentor
Just to confirm, would it be accurate to make the following statement—

“Whenever t & B are rational, A is irrational. Whenever A & B are rational, t is irrational.”

?
Are you referring to this equation?
##A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)} ##
If so, t could be rational but ##\cos(t)## and ##\sin(t)## (e.g., ##\sin(1)## or ##\cos(1)##) definitely don't have to be rational. The equation above is complicated enough that I don't see your conclusions being valid.

• Devin-M
Mentor
The reason I ask is the 1st paragraph in this paper:

https://aapt.scitation.org/doi/abs/10.1119/1.11441?journalCode=ajp
This paragraph?
It is shown that, in general, there is no solution to the title problem. However, there is always a smooth curve, that is not a solution of the Euler-Lagrange equation, along which the time of travel is arbitrarily close to the minimal, which is the time of travel along the curve that satisfies the Euler-Lagrange equation but not all boundary conditions.

Devin-M
Yes.

Mentor
I don't know what that paragraph has to do with your statement about combinations of A, B, and t being rational or irrational in the equation of post #17.

Devin-M
If we know A, and it’s rational and in meters, we know the gravitational potential difference and therefore velocity at t from the untruncated cusp. If B is rational as well, and so is t, and from t we find r, wouldn’t it mean there is a solution?

Mentor
If we know A, and it’s rational and in meters, we know the gravitational potential difference and therefore velocity at t from the untruncated cusp. If B is rational as well, and so is t, and from t we find r, wouldn’t it mean there is a solution?
I'm confused now. The two equations you've given in this thread are:
##A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)} ##
If A and B are known, it is very likely not possible to solve for t analytically. The equation is complicated enough that the Lambert-W function won't be helpful, I believe.
##r= \frac B {2[(\pi-t) + \sin(t)]}##
The latter equation you described as being part of the solution
In the second equation above, if r and B are known, it is also very doubtful that one can solve for t analytically.

The fact that this or that variable is rational or irrational doesn't enter into things at all. The main difficulty is that you have the variable t appearing by itself and also as an argument to one or both of the sine and cosine functions.

That is the difficulty, not whether something or other is rational.

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If we know A, and it’s rational and in meters, we know the gravitational potential difference and therefore velocity at t from the untruncated cusp. If B is rational as well, and so is t, and from t we find r, wouldn’t it mean there is a solution?
I think you are confusing an analytical solution with a solution. Just because we cannot write down the solution in terms of analytical functions doesn't mean that there isn't a solution. As I stated earlier, your original problem: $$A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)}$$, solving for t when A=-0.077 and B=1.0, has a solution. In fact it has 6 solutions. You can calculate those solutions to as many digits as you like.

Devin-M
So does it take a finite time to calculate the quickest route between 2 horizontal points w/ a given initial velocity (the generating radius (r) , initial (t) and final theta (2pi-t) of a portion of a cycloid)?

Mentor
So does it take a finite time to calculate the quickest route between 2 horizontal points w/ a given initial velocity (the generating radius (r) , initial (t) and final theta (2pi-t) of a portion of a cycloid)?
What does this question have to do with anything? I can calculate a numeric solution to a specific number of decimal places in X seconds, and a computer program can do the same in less time, but so what?

• Devin-M
Devin-M
So if A is -1 and B is 1, does t have an infinite or finite # of non repeating decimals?

Mentor
So if A is -1 and B is 1, does t have an infinite or finite # of non repeating decimals?
t is almost certainly irrational, but so what? The same would be true for most irrational choices of A and B.

• Devin-M
Devin-M
So if the initial velocity is 4m/s in 8m/s^2 gravity, then A is exactly -1m. Suppose I want to build a ramp to go 1 meter (B) horizontally on the quickest path (optimized for time) with 4m/s initial velocity in 8m/s^2 gravity. To find the quickest route we need to know the cycloid generating radius (r) calculated from (B) and (t) but to find this requires first doing a possibly endless task (finding t). Is it possible to find r and therefore build the ramp which has the quickest route?

So does it take a finite time to calculate the quickest route between 2 horizontal points w/ a given initial velocity (the generating radius (r) , initial (t) and final theta (2pi-t) of a portion of a cycloid)?
If you want a finite accuracy, yes.

• Devin-M
Like @Mark44 , I don't understand where you are coming from. It is easy to solve this type of problem numerically, and a computer will find the answer to any reasonable degree of accuracy as soon as you hit 'return'. Whether the answer is rational, irrational, or can be expressed analytically is basically irrelevant.

• Devin-M
Devin-M
I was just wondering if after any amount of time the computer program would be done solving for t, having found it exactly. I think you are saying the answer is that it’s an endless task though I don’t know for sure.

Yes, it's an endless task and the answer will never be known exactly. Just like you will never know the value of pi exactly.

• Devin-M
Homework Helper
Gold Member
It would be possible to determine the infinite Taylor series of the solution or some other algorithm. After that, the length of the task depends on the accuracy that you require. If you are asking for a closed-form solution, those are not as common in real applications as you might think.

• Devin-M
Devin-M
So then does it become possible to show, at a finite degree of accuracy, that in 9.8m/s^2 gravity, with 1m/s initial velocity that the minimum travel time between 2 horizontal points separated horizontally by 1 meter is 0.602... seconds via:

##Time = sqrt(r/g) * dt## Last edited: