I Is it possible to solve for “t?”

[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?

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.

 

[phyzguy]

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]

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.

 

[phyzguy]

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.

 

[FactChecker]

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]

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:

$Time = sqrt(r/g) * dt$

 

[pbuk]

1. You have confused things greatly by putting $t$ in your equations to represent an angle which you have called theta. It would be much clearer if you had used $\theta$ (i.e. ## \theta ## to mean theta leaving $t$ for time.
2. Where did you get $Time = sqrt(r/g) * dt$ from? What does it mean? Does $dt$ here have something to do with time?
3. How do you know that your assumption that the optimal curve when the initial velocity is non-zero is a cycloid is correct?
4. You have not specified the final velocity anywhere, can it be zero?
5. You start talking about depth and elsewhere you mention a distance of 5,000 metres. Do you realize that gravity decreases with depth? (edit: oops, no it doesn't!)
6. Do you realize that friction, and particularly air resistance, becomes very significant at increased velocity, meaning that for any real application you need to provide additional power to avoid falling short of the destination and falling back down the slope?

You have asked a lot of questions that have been difficult to interpret, and then it turns out that the question you have asked is not really relevant. Why don't you ask what it is you actually want to know and then we can work out which details are relevant?

 

[mfb]

Do you realize that gravity decreases with depth?

It increases a bit (PREM) until you reach the core.

 

[pbuk]

It increases a bit (PREM) until you reach the core.

Good catch, thanks - edited.

 

[Devin-M]

Where did you get Time=sqrt(r/g)∗dt from? What does it mean? Does dt here have something to do with time?

I got that one from here:
https://mathworld.wolfram.com/TautochroneProblem.html

 

[Devin-M]

How do you know that your assumption that the optimal curve when the initial velocity is non-zero is a cycloid is correct?

I got that one from pg 116:
http://classicalmechanics.lib.rochester.edu/pdf/vpcm.pdf

 

[pbuk]

Where did you get Time=sqrt(r/g)∗dt from? What does it mean? Does dt here have something to do with time?

I got that one from here:
https://mathworld.wolfram.com/TautochroneProblem.html

Are you referring to Equation 12 $dt = \sqrt { \dfrac a g } d \theta$? How are you calculating the parameter a (or r in your equation) and the initial angle $\theta_0$?

 

[pbuk]

If you show your workings then there would be something to check, a table of numbers is not much help.

 

[pbuk]

How are you calculating ... the initial angle $\theta$?

From your table it looks like you are guessing it. You don't have to do that, the initial point is the vertical distance below the cusp where the difference in potential energy equals the initial kinetic energy.

 

[Devin-M]

I calculated t (initial angle) and r (generating radius or a) using some logic steps:

https://u.pcloud.link/publink/show?code=kZq8VHXZIECHjF10TLJpQhMs7l1WEROKqDt7

For t, the algorithm first guesses 1 radian, runs it through the A= formula, compared it to the desired depth to determine if 1 rad is too big or small. Then it determines an increment and continues refining until A= is as close to the desired depth as possible after a certain # of iterations...

 

[Devin-M]

Are you referring to Equation 12 dt=a/gdθ? How are you calculating the parameter a (or r in your equation) and the initial angle θ?

The other formulas are:

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...

The other helpful part of the solution is:

r=B/(2((pi−t)+sin(t)))

Where r is the cycloid generating radius

(the lowercase a from your quote equals the lowercase r)

(uppercase A does not equal the lower case a)

 

[pbuk]

OK, I see what you are doing. The calculations in the sheet you linked look OK, you could improve this by using the solver to make 'Freefall Equivalent Error (m)' = 0 by adjusting 'Cycloid Initial Theta Best Guess (rad)'.

Bear in mind that the assumption is that the initial velocity is in the direction of the curve (which is pretty steep), not horizontal.

 

[Devin-M]

you could improve this by using the solver to make 'Freefall Equivalent Error (m)' = 0 by adjusting 'Cycloid Initial Theta Best Guess (rad)'

& therein lies the endless task...

 

[pbuk]

& therein lies the endless task...

Its not an endless task, its an iterative task, and the solver in Excel will do it for you in a millisecond.

 

[Devin-M]

Its not an endless task, its an iterative task, and the solver in Excel will do it for you in a millisecond.

hmm... i thought there was no exact solution for t when A and B are both rational... (or t is irrational)

I was just wondering if after any amount of time the computer program would be done solving for t, having found it exactly.

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.

 

[FactChecker]

Its not an endless task, its an iterative task, and the solver in Excel will do it for you in a millisecond.

Excel will have some error tolerance number in the algorithm to tell it when the answer is close enough to stop. Whether you want to call this "endless" or not depends on whether Excel's tolerance is good enough for your application. In any case, it is not the same as a closed formula. Even with a closed formula, the value that you get from a computer is often an approximated evaluation of the formula which may be obtained iteratively.