Logarithmic spiral

[jpi]

Hi

I have some questions regarding a puzzle.

This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.

If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?

And how do i find the length ?

Regards
Jacob

 

[galactus]

\(\displaystyle r=a\cdot e^{b{\theta}}\) is the straight line distance from the origin to a point on the curve.

Perhaps you can use the given info to find a and b. They are arbitrary constants that have to do with how tightly the spiral is wound and so forth.

The arc length is given by \(\displaystyle s=\frac{a\sqrt{1+b^{2}}e^{b{\theta}}}{b}\), which can be found by doing a search.

You are given two distances. A at \(\displaystyle {\theta}={\pi}\) and B at \(\displaystyle {\theta}=2{\pi}\). Starpunkt is at \(\displaystyle {\theta}=0\)

There is lots on google search regarding these.

 

[wjm11]

This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.

If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?

To expand on Galactus’ advice:

Let:
R1 = distance from origin to startpunkt
R2 = distance from origin to A
R3 = distance from origin to B

Use r = ae^(b*theta) to create appropriate expressions for R1, R2, and R3.

Substitute those expressions into the following equations:

Absolute value(R1) + Abs(R2) = 86.23 or just R1 + R2 = 86.23

Absolute value(R3) - Abs(R1) = 75.41 or just R3 – R1 = 75.41

At this point, you have two equations and two unknowns. Hope that helps.

Edited to correct sign error.

 

[jpi]

Absolute value(R1) + Abs(R2) = 86.23 or just R1 – R2 = 86.23

Shouldn't it be +on both ?

 

[wjm11]

Shouldn't it be +on both ?

You're right, of course. I was erroneously thinking you might get a negative value for the distance to A. Good catch.

 

[jpi]

OK

It a long time since i have done these kind of calculations :?
Is this correct so far ?

\(\displaystyle a+ae^{b\pi} = 86,23\) and \(\displaystyle ae^{2b\pi}-a = 75,41\)

isolating a:
\(\displaystyle ae^{2b\pi}-a = 75,41\)

\(\displaystyle ae^{2b\pi} = 75,41+a\)

\(\displaystyle e^{2b\pi} = \frac{75,41+a}{a}\)

\(\displaystyle e^{2b\pi} = \frac{75,41}{a}+1\)

\(\displaystyle e^{2b\pi}-1 = \frac{75,41}{a}\)

\(\displaystyle \frac{1}{e^{2b\pi}-1} = \frac{a}{75,41}\)

\(\displaystyle 75,41*{\frac{1}{e^{2b\pi}-1}} = a\)

 

[wjm11]

Your work looks fine, but I'd approach differently: factor a out of the two terms in both equations and solve for each equation in terms of a. Next set the equations equal to each other and solve for b.

 

[jpi]

Your work looks fine, but I'd approach differently: factor a out of the two terms in both equations and solve for each equation in terms of a. Next set the equations equal to each other and solve for b.

Sorry, English is not my primary language and i don't understand what you mean

I thought of doing the same with the other equation (86,23=...) and then i have two equations with a= something and then set then equal to one another.

A Differnt approach: Take the equation i have solved for a and put the result in a's place in the other equation. Then solve for b. Then put b in one of the equations and i should get a, . Correct ? Actually i should get the same value for a in both equations

 

[jpi]

Just to be sure you know : theta=0 at Startpunkt

 

[galactus]

Yes, I made a mistake and just fixed it in my original post.

 

[wjm11]

I thought of doing the same with the other equation (86,23=...) and then i have two equations with a= something and then set then equal to one another.

Yes, that is correct. That is what I was suggesting. That will work just fine. (By the way, I think your English is excellent. I was guessing you might be European by your use of a "," where we use a "." or "decimal point" in a number.)

 

[jpi]

I have now solved the puzzle, and got the answer correct !!!

Thanks a lot. It is higly appreciated

/Jacob

 

[galactus]

Very good. I saw you were on the right track with your last post. Good work.

Log spirals are very interesting. There's lots to learn to you google it.