Is it possible to solve for “t?”

[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.

 

[pbuk]

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

It's got nothing to do with being irrational, it is simply not helpful to talk about this task as 'endless'.

It is true that it is not possible to write down an 'exact' solution to the problem, but neither is it possible to write down the 'exact' value of 1/3 (a rational number) in the decimal number system, would you refer to that as endless?

There are things in mathematics which are 'endless', such as searching for the limit of the sum $1 + \frac12 + \frac13 + \frac 14 ...$, but that is not the situation we have here.

 

[Mark44]

It is true that it is not possible to write down an 'exact' solution to the problem, but neither is it possible to write down the 'exact' value of 1/3 (a rational number) in the decimal number system, would you refer to that as endless?

Excellent point that seems to get to the heart of what the OP is asking about.
I can solve the equation $3x = 1$ algebraically to get the exact answer, but it will take an "endless" number of digits to write the exact value as a decimal fraction.

But so what? Rational or irrational makes just about zero difference here.

My suspicion is that what the OP is trying to get at is that the object can slide down the tube in a finite amount of time, but writing all of the decimal digits would take an infinite amount of time. I could be wrong, but if not, this is not something to be concerned about.

 

[Devin-M]

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...

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.”

?

I suppose I thought that, when A & B are rational (each can be expressed as a ratio between integers) that, since the formula can’t be rearranged algebraically to solve for t, that t might not be able to be expressed as a ratio between integers and hence t would be “irrational” though I’m not sure.

I was imagining, if t is irrational, it leading to a scenario where someone with a fast computer solves t to very many digits, and from t and B calculates r, and from r, g and t, calculates the travel time and says “this is the quickest route.” But then someone else calculates to more digits, and finds an even faster time. Could this lead to a scenario where one can’t find the “quickest” route because someone with a faster computer and more time can always calculate t to more digits than the person before (more digits of t affecting and altering the proposed “quickest” cycloid generating radius), so as time goes on ever faster routes can be found ad nauseum?

 

[physlosopher]

I was imagining, if t is irrational, it leading to a scenario where someone with a fast computer solves t to very many digits, and from t and B calculates r, and from r, g and t, calculates the travel time and says “this is the quickest route.” But then someone else calculates to more digits, and finds an even faster time. Could this lead to a scenario where one can’t find the “quickest” route because someone with a faster computer and more time can always calculate t to more digits than the person before (more digits of t affecting and altering the proposed “quickest” cycloid generating radius), so as time goes on ever faster routes can be found ad nauseum?

So, two points:

(1) I think that what others are trying to say is that even if the "exact solution" for t is a number with infinitely many digits, the problem you seem to be worried about holds for any such number. For example, maybe I know the answer is 1/3 (a rational number). If I have to write that out as a decimal, I know exactly how to do that - start with 0.33 and keep adding 3's on the right. So it would take me an infinite amount of time to write it down exactly, but not nearly an infinite amount of time to write it out to a reasonable degree of precision.

Apply the same reasoning to a number like $\pi$ (an irrational number). If you give me a radius, I can calculate the circumference $2\pi r$ of a circle to arbitrarily many digits, depending on how many digits of $\pi$ I want to compute and use. No, it will never have infinitely many digits, but there's never going to be an application in which you need to know infinitely many digits. Knowing the value of the circumference that's more precise than any of your measuring devices, for example, is more than enough. More digits than that become redundant unless you're going to build a better tape measure.

(2) Both in the circle example and in your example of an extremal trajectory, using more digits will not radically change your final answer. Rather, you should expect that if you keep computing with more and more precision, your circumference or trajectory will be converging to some particular thing. No, you won't be able to say what it is with infinitely many digits, but you'll have, for example, a trajectory that changes so little as you introduce more digits for t that you very quickly stop being able to measure the resulting differences in travel time. Think of computing with t to 10 digits, and then with 15, and getting two trajectories. They will probably be similar to the point that a stopwatch could not measure the difference in time between taking each path, and indeed they're probably similar enough that no instrument could chart a course that knows how to distinguish between them. Furthermore, a computer would take very little time to do either of these calculations.

In fact, you'll find that you waste much more time trying to calculate to infinite digits than to just travel on the trajectory that's good to 10 digits, even while your measurable time spent on each of these paths would not be different.

Am I understanding correctly that you think there's some paradox in taking infinite time to calculate a trajectory that minimizes time? I'm not sure whether you just find this kind of ironic or whether it really bothers you mathematically. Hope this helps!

 

[pbuk]

I suppose I thought that, when A & B are rational (each can be expressed as a ratio between integers) that, since the formula can’t be rearranged algebraically to solve for t, that t might not be able to be expressed as a ratio between integers and hence t would be “irrational” though I’m not sure.

Nearly all numbers are irrational but they don't seem to worry about it and nor should you

I was imagining, if t is irrational, it leading to a scenario where someone with a fast computer solves t to very many digits, and from t and B calculates r, and from r, g and t, calculates the travel time and says “this is the quickest route.” But then someone else calculates to more digits, and finds an even faster time. Could this lead to a scenario where one can’t find the “quickest” route because someone with a faster computer and more time can always calculate t to more digits than the person before (more digits of t affecting and altering the proposed “quickest” cycloid generating radius), so as time goes on ever faster routes can be found ad nauseum?

No, there is only one fastest route.

Bear in mind that you have used the approximate value of 9.8 m/s² for g: there is no point specifying your solution to more than 3 significant figures because it will change if use use a different value of g (note that g varies by about 0.05 m/s² between the equator and the poles and there are also local variations due to altitude and geophysical anomalies).

 

[WWGD]

I assume the Implicit Funcion Theorem could give a definitive answer on the possibility of isolating t?

 

[pbuk]

Not my area of expertise but won't the Implicit Function Theorem just give us existence not construction? We don't have a problem with existence - we don't even have a problem with calculation; this is simple root finding for a (in the region of interest) monotonic function so we can achieve arbitrary precision in logarithmic time.

 

[Mark44]

I assume the Implicit Funcion Theorem could give a definitive answer on the possibility of isolating t?

I don't think so. I agree with @pbuk that this theorem just talks about the existence of a function. From the wiki page, https://en.wikipedia.org/wiki/Implicit_function_theorem

The implicit function theorem gives a sufficient condition to ensure that there is such a function.

Not my area of expertise but won't the Implicit Function Theorem just give us existence not construction?

That's how it seems to me.

 

[WWGD]

I don't think so. I agree with @pbuk that this theorem just talks about the existence of a function. From the wiki page, https://en.wikipedia.org/wiki/Implicit_function_theorem

That's how it seems to me.

Agreed, but maybe a 'no', not possible to represent even locally, would shed some light. Let me look at the theorem again see if I can find a constructive corollary of some sort.

 

[pbuk]

It is probably worth a summary as there have been a few diversions and dead ends on the way.

We are trying to find the shortest journey time between points $A$ and $B$ at the same elevation for an object under the influence of gravity constrained without friction to a path. The object has initial velocity of $v_0$ in the initial direction of travel, and arrives at B with equal speed.

By other work referenced, we know that the solution is part of a cycloid defined by
$$ \begin{align*} x &= r (t - \sin t) \tag 1 \\
y &= r (1 - \cos t) \tag 2 \end{align*} $$
at point $A$ we must have
$$ mgy_A = 0.5mv_0^2 \implies y_A = \frac{v_0^2}{2g} \tag 3 $$
and we know the distance $X$ between $A$ and $B$ and that $x_A = x_B$ so
$$ 2 x_A + X = 2 \pi r \implies X = 2 (\pi r - x_A) \tag 4 $$

We now have two unknowns, $t_0$ (the value of the parameter t at X) and $r$ (the generating radius). There are a number of ways of proceding: the OP (if I remember rightly) has estimated $t_0$ and used (2) and (3) to find the corresponding $r$; substituted this into (1) and (4) to find the resulting $X$ and reduced the error by bisection.

 

[Devin-M]

2 comments...

By other work referenced, we know that the solution is part of a cycloid defined by

I believe that equations 1 & 2 specify a cycloid that proceeds above the x-axis from the origin, but the needed path is an upside-down or inverted cycloid that proceeds below the x-axis from the origin.

we know the distance X between A and B and that xA=xB so

I believe you meant yA=yB instead of xA=xB

 

[Devin-M]

If I solve for t and r to 5 decimal places in 9.8m/s^2 gravity where the horizontal distance between the start and endpoint is 1 meter and the initial velocity is 1m/s I get the following parametric graph:

Parametric Graph: https://www.desmos.com/calculator/0k2dfjrnbw

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

...& travel time of 0.60214 seconds.

 

[Devin-M]

and we know the distance X between A and B and that xA=xB so
(4)2xA+X=2πr⟹X=2(πr−xA)

We now have two unknowns, t0 (the value of the parameter t at X)

There may be an error in equation 4... I think the equation is true for a complete cycloid when you know the horizontal distance between the cusps (except 2xA should simply be xA), but isn’t true when the cycloid is truncated by a certain distance in the y-axis when you know the horizontal distance between the truncated cusps.

I believe xA + X = 2pi * r only when the cycloid hasn’t been trimmed in the y-axis by specifying a range for t other than 0<=t<=2pi

 

[pbuk]

I believe that equations 1 & 2 specify a cycloid that proceeds above the x-axis from the origin, but the needed path is an upside-down or inverted cycloid that proceeds below the x-axis from the origin.

You can flip the sign of equation (2) if you want and work with negative values of $y$ (and g) everywhere, but the maths is the same if we keep equation (2) as it is and measure $y$ downwards (in other words flip your drawing upside down).

I believe you meant yA=yB instead of xA=xB

I had the right picture in my head and the conclusion is right but of course $x_A \ne x_B$. The workings should have been $$ x_B = x_A + X = 2 \pi r - x_A \implies X = 2 (\pi r - x_A) \tag 4 $$

 

[pbuk]

There may be an error in equation 4... I think the equation is true for a complete cycloid when you know the horizontal distance between the cusps (except 2xA should simply be xA), but isn’t true when the cycloid is truncated by a certain distance in the y-axis when you know the horizontal distance between the truncated cusps.

I believe xA + X = 2pi * r only when the cycloid hasn’t been trimmed in the y-axis by specifying a range for t other than 0<=t<=2pi

View attachment 273021

No, for a complete cycloid the horizontal distance is simply $2 \pi r$.

Setting $x = x_A, t = t_0$ in equation (1) and substituting it into equation (4) you get
$$ X = 2 (\pi r - r (t_0 - \sin t_0)) \implies r = \frac{X}{2(\pi - t_0 + \sin t_0)} $$ which is exactly what you have in cell C14: =(C3/(2*(PI()-C162+SIN(C162)))).

 

[etotheipi]

Apologies if I missed this, but how did we prove that the optimal path is a cycloid? When I tried to solve the variational problem, I ended up with a differential equation$$y'^2 y''(v_0^2 - 2gy) = (1+y'^2)((v_0^2 - 2gy)y'' - g)$$that I don't know how to solve. Although I can try the cycloid ansatz and see if it works

 

[pbuk]

Apologies if I missed this, but how did we prove that the optimal path is a cycloid?

That was my first question too, but if you think about it then it must be so because all we are doing by giving the particle an initial velocity is considering its position at some point along the solution to a different Brachistochrone problem with a higher stationary origin. The potential energy difference at that origin must equal the new kinetic energy.

Also the OP gave a reference:

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

Edit: this does of course assume familiarity with the classic Brachistochrone problem and it's cycloid solution as set out by Bernoulli and by Newton.

 

[etotheipi]

Thanks, interesting problem! As a sanity check, I let $v_0 = 0$ in my differential equation and then trialled$$(x,y)= \left( a(\theta - \sin{\theta}), a(1-\cos{\theta}) \right), \quad y' = \frac{\sin{\theta}}{1-\cos{\theta}},\quad y'' = \frac{-1}{a(1-\cos^2{\theta})}$$ and that satisfies the equation. For non-zero $v_0$, I think it will be solved by something of the form $(x,y)= \left( a(\theta - \sin{\theta}) - b, a(1-\cos{\theta}) - c \right)$

 

[pbuk]

For non-zero $v_0$, I think it will be solved by something of the form $(x,y)= \left( a(\theta - \sin{\theta}) - b, a(1-\cos{\theta}) - c \right)$

You have three unknowns there (a, b, c) but only 2 constraints (initial velocity and distance). So the correct equations are (1) and (2) in my post a few up.

 

[etotheipi]

You have three unknowns there (a, b, c) but only 2 constraints (initial velocity and distance). So the correct equations are (1) and (2) in my post a few up.

But I think $c$ is functionally dependent upon $b$, in order to constrain that trajectory intersects the origin.