Is it possible to solve for “t?”

[Devin-M]

TL;DR

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?

For reference: https://www.physicsforums.com/threads/brachistochrone-problem-w-initial-velocity.995659/

 

[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.

 

[Mark44]

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]

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...

 

[phyzguy]

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!

 

[mfb]

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:
$A = \frac{-B}{2}\cdot\frac{1-cos(t)}{ \pi-t+sin(t)}$

 

[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]

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.”

 

[FactChecker]

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?

 

[FactChecker]

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.”

?

 

[Mark44]

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]

The reason I ask is the 1st paragraph in this paper:

https://aapt.scitation.org/doi/abs/10.1119/1.11441?journalCode=ajp

 

[Mark44]

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.

 

[Mark44]

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?

 

[Mark44]

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.

 

[phyzguy]

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)?

 

[Mark44]

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]

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

 

[Mark44]

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.

What's your point here?

 

[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?

 

[phyzguy]

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.