Calculate distance between ends of a circle segment

[solvejskovlund]

I was playing with a small piece of rope I had on the table, moved it to a curve - circle segment - and figured I wanted to calculate the distance between ends of the curve - and failed. Now I'm confused and embarrassed.

Here is what I did:
The rope has length L. I put the ends of the rope on a ruler (straight line). The ends are free to move along that line, but not allowed to leave that line.
Now I move the center point of the rope L/5 away from the ruler. Given the restrictions that the center of the rope is fixed L/5 away from the ruler, and the ends has to be on the ruler line I arrange the rope into a part of a circle.
How can the distance between the ends of the rope be calculated?

 

[Lnewqban]

Did I understand your description?

 

[solvejskovlund]

Yes

And the arc has a constant radius. (It is part of a circle)

 

[Lnewqban]

Yes

And the arc has a constant radius. (It is part of a circle)

What have you tried to determine that radius so far?

This may help:
https://en.wikipedia.org/wiki/Circular_segment

 

[solvejskovlund]

I couldn't figure a way to calculate chord (c) without finding the radius (R). Determining the radius turned out to be where I failed in my logic thinking. Thanks to the wiki formulas, now I've also failed using math!

Looking at the formulas in the wiki, there is
R = h/2 + c²/8h
h in my case is known to be L/5. To find c, I need R.

Next;
c = 2 * sqrt( R² - ( R - h )²) = 2 * sqrt( 2Rh - h² )
h in my case is known to be L/5. To find c, I need R.

Using this to find c²:
c² = 4 * ( 2Rh - h²) = 8Rh - 4h²

Setting the later into the first, I get:
R = h/2 + c²/8h
R = h/2 + ( 8Rh - 4h² ) /8h
R = h/2 + 8Rh/8h - 4h² /8h
R = h/2 + R - h/2
R - R = h/2 - h/2
0 = 0
This only tells me that if the rope initially was of length 0, everything else would also be 0. My rope was longer than 0.

This is the obvious result because the two formulas used is determined from each other.

I think the main issue here is that there is a need for one more known parameter in the calculations. As the rope is there as a proof, there must be one more known parameter that I overlook.
I know the length of the rope: L
I know the height of the arc: h=L/5
I know radius is constant

Somehow this information should be used to determine how large part of a circle the arc is.
If the arc had been a half circle, then c=2R and R=h and L=π*R = π*h => h=L/π. That's just logic. There must be some math behind that as well - math that works when h=L/x where x > π, causing h<R.
The known extra parameter here was that the arc was half a circle.

 

[erobz]

This doesn't seem to be a trivial problem now that I look at it...I would say it probably ( i.e. almost definitely-unless I'm missing something) requires a numerical solution. However, that shouldn't stop us from trying...

You have a few equations to work with involving $\theta,L,R, x$ ( we only know $L$ and the fact is forms part of a circular arc )

Hint: Think arclength (circular sector), tangent, and Pythagoras ( red triangle). Can you write them for us?

Please see LaTeX Guide to properly format your developed equations.

 

[solvejskovlund]

Some formulas I see related:

$\frac{L}{2πR} = \frac{2θ}{2π} = \frac{θ}{π}$ (segment length vs circle length = angle vs circle angle)

R² = x² + (R-L/5)² (pythagoras)

$x=R*\sin(\frac{\Theta}{2})$

Combining these, knowing that x=2c, and h=L/5, I just get the formulas of the wiki as stated in above post.

I still think there is one little information that I'm missing out from looking at the drawing.

 

[docnet]

hint: you can get an expression for $R-h$ in terms of $R$ and $\theta$ using basic trigonometry.

 

[solvejskovlund]

R-h=R*cos(θ)
$cos(θ)=\frac{R}{R}-\frac{h}{R}=1-\frac{L}{5R}$
Could also be written as
$\frac{L}{5R}=1-cos(θ)$
$\frac{L}{R}=5-5*cos(θ)$

From before I have:
$\frac{L}{2πR} = \frac{2θ}{2π} = \frac{θ}{π}$ (segment length vs circle length = angle vs circle angle)
$\frac{L}{R} = \frac{2πθ}{π} = 2θ$

Combining those:
$\frac{L}{R}=5-5*cos(θ) = 2θ$
At least that's an equation with only one unknown. But I don't see how to move on from there.

 

[erobz]

Combining those:
$\frac{L}{R}=5-5*cos(θ) = 2θ$
At least that's an equation with only one unknown. But I don't see how to move on from there.

I'll redefine the angle to be $\theta$ ( as opposed to $\theta/2$ - it's better that way. )

So we get:

$$ L = R 2\theta \tag{1} $$

$$ \frac{R - \frac{L}{5}}{R} = \cos \theta \tag{2} $$

Combining those I get what you do

$$ 5 \left( 1 - \cos \theta \right) = 2 \theta \tag{3} $$

$\text{3}$ is unfortunately the numerical solution that needs to be found.

However, you can continue to work on $x$ with another trig relationship, and Pythagoras. You are going to eliminate $R$ in these steps between two equations.

 

[solvejskovlund]

Uh. I didn't realize I had left out the /2 when using θ. Glad you pointed that out.

I don't see which trig relationship you're pointing to.

 

[gmax137]

Combining those I get what you do

$5(1−cos⁡θ)=2θ$

3 is unfortunately the numerical solution that needs to be found.

You can solve graphically. Or rewrite it as
$2θ-5(1−cos⁡θ)$
And then use Excel "goal seek" to make this zero by varying theta
$$
\begin{array}{|c|c|c|c|}
\hline theta&5(1-cos(theta))&2theta&difference\\
\hline 0.84993679&1.699846834&1.69987358&2.67453E-05 \\
\hline
\end{array}$$

 

[Lnewqban]

Could a mathematical solution be found for the intersections of a line, an arc and a cycloid?

 

[Lnewqban]

Just to be used as a reference, I have graphically found the following proportions:

 

[erobz]

You can solve graphically. Or rewrite it as
$2θ-5(1−cos⁡θ)$
And then use Excel "goal seek" to make this zero by varying theta
$$
\begin{array}{|c|c|c|c|}
\hline theta&5(1-cos(theta))&2theta&difference\\
\hline 0.84993679&1.699846834&1.69987358&2.67453E-05 \\
\hline
\end{array}$$

I just was trying to verify the OP understood the hopes of analytical solution for $x$ are slim because of the nature of this equation.

 

[erobz]

I don't see which trig relationship you're pointing to.

You've already found it.

$$ x = R \sin \theta \tag{4} $$

and Pythagoras

$$ x^2 + \left( R-\frac{L}{5} \right)^2 = R^2 \tag{5} $$

Use these to eliminate $R$ in Eq.$\text{4}$.

You should be left with a quadratic in $x$, treating $\sin \theta$ as some constant, but knowing said constant requires the solution of $\text{3}$ for $\theta$.

 

[solvejskovlund]

Thought you ment something that wasn't already listed.
(4) gives
$R=\frac{x}{sin(\Theta)}$

(5) gives
$x^2 + R^2-2R\frac{L}{5}+\frac{L^2}{25}=R^2$
$x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0$

combined:
$x^2 -2* \frac{x}{sin(\Theta)} *\frac{L}{5}+\frac{L^2}{25}=0$
Having x and θ as unknown isn't any better than having R and θ as unknown. Combine with some other eq, this could be changed to x and R as unknown.... Won't help much either.

I though this task would be an easy one. Turns out it wasn't. If h=L/π it turns out to be very easy, because there is one more known parameter. From a calculation point of view, what exactly is the known parameter when h=L/π, that is not known when i.e. h=L/5 (that is h<L/π) ?
Edit: ah. that is both x=R and θ=π/2

 

[erobz]

Thought you ment something that wasn't already listed.
(4) gives
$R=\frac{x}{sin(\Theta)}$

(5) gives
$x^2 + R^2-2R\frac{L}{5}+\frac{L^2}{25}=R^2$
$x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0$

combined:
$x^2 -2* \frac{x}{sin(\Theta)} *\frac{L}{5}+\frac{L^2}{25}=0$
Having x and θ as unknown isn't any better than having R and θ as unknown. Combine with some other eq, this could be changed to x and R as unknown.... Won't help much either.

I though this task would be an easy one. Turns out it wasn't. If h=L/π it turns out to be very easy, because there is one more known parameter. From a calculation point of view, what exactly is the known parameter when h=L/π, that is not known when i.e. h=L/5 (that is h<L/π) ?
Edit: ah. that is both x=R and θ=π/2

I think you are missing what I'm saying. Solve (5) for $R$, and sub it into (4). $\theta$ is not unknown anymore. It is found by solving (3) independently for $\theta$ in theory (numerically) of course - you are (likely) not going to find the analytic expression unless the mathematicians have a trick for this one. After doing so, you are left with an equation of the form:

$$a x^2 + b x + c = 0 $$ which has the solution:

$$ x = \frac{-b \pm \sqrt{ b^2 - 4 a c }}{2a} $$

That gets you $x$.

The system is fully constrained pending you find $\theta$ as @gmax137 suggest in #12 or by some other numerical technique. Your combined equation is correct , but you need to solve $\text{3}$ to get any further. The result of solving (3) becomes a part of $b$ in the quadratic.

 

[erobz]

How do you delete post... I can't find it anymore! Please delete this when someone gets a chance. Thank You.

 

[solvejskovlund]

For (5) I had
$x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0$

Solving (5) for R:
$2R\frac{L}{5}=x^2+\frac{L^2}{25}$
$R=x^2\frac{5}{2L}+\frac{L^2}{25}*\frac{5}{2L}$
$R=\frac{5x^2}{2L}+\frac{L}{10}$

Insert into (4)
$x=(\frac{5x^2}{2L}+\frac{L}{10})*sin(\Theta)$
$x^2*\frac{5*sin(\Theta)}{2L}-x+\frac{L*sin(\Theta)}{10}=0$
$x=\frac{1\pm \sqrt{1-4*\frac{5*sin(\Theta)}{2L}*\frac{L*sin(\Theta)}{10}}}{2\frac{5*sin(\Theta)}{2L}}$
$x=\frac{1\pm \sqrt{1-sin(\Theta)^2}}{\frac{5*sin(\Theta)}{L}}$

I don't see the point of finding more ways to express relations between the variables using the formulas listed. It is shown that this task could not be solved without going the route of trial and error, as shown in post #12.

 

[erobz]

For (5) I had
$x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0$

Solving (5) for R:
$2R\frac{L}{5}=x^2+\frac{L^2}{25}$
$R=x^2\frac{5}{2L}+\frac{L^2}{25}*\frac{5}{2L}$
$R=\frac{5x^2}{2L}+\frac{L}{10}$

Insert into (4)
$x=(\frac{5x^2}{2L}+\frac{L}{10})*sin(\Theta)$
$x^2*\frac{5*sin(\Theta)}{2L}-x+\frac{L*sin(\Theta)}{10}=0$
$x=\frac{1\pm \sqrt{1-4*\frac{5*sin(\Theta)}{2L}*\frac{L*sin(\Theta)}{10}}}{2\frac{5*sin(\Theta)}{2L}}$
$x=\frac{1\pm \sqrt{1-sin(\Theta)^2}}{\frac{5*sin(\Theta)}{L}}$

I don't see the point of finding more ways to express relations between the variables using the formulas listed. It is shown that this task could not be solved without going the route of trial and error, as shown in post #12.

There is some kind of communication breakdown, because I thought you wanted the chord length $2x$. There it is. Find an analytic solution to (3) and you have the analytic formula you seek!

Has it been proven that no such solution exists (or could possibly exist) for (3) - or did humanity just give up too soon? Maybe some new mathematics need invented.

Even if you don't succeed in this goal in your lifetime, there is still a benefit to having the final result for $x$ algebraically. You plug in the value you calculate for $\theta$ and out comes the answers you seek.

 

[solvejskovlund]

A unknown x given by another unknown θ that is derived from x, doesn't solve anything. I'm surprised this wasn't straight forward math.

 

[erobz]

A unknown x given by another unknown θ that is derived from x, doesn't solve anything. I'm surprised this wasn't straight forward math.

The angle is fixed independent of $x$. There is no $x$ in the solution for the subtended angle of the arc. If $\theta$ wasn’t buried in a trig function it would have an analytic solution, meaning you could write a formula down for it.

$\theta$ is not unknown. It’s just buried in an equation that is difficult to write a formula down for. It's not an explicit solution, but it's a solution none the less. When you say things like what is quoted it's hard to tell if you are still needing clarification on this point.

I get it...it's frustrating! I'm going to go out on a limb and say it's the things you can write down formulaically are what is rare, so you better get used to it.