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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I made this diagram in a free CAD program.

Can anyone recognize the curve as anything??

I made it and I don't know the answer yet myself.

Is this curve a well-known curve??

Can you guess how I made it???

With what criteria??

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It looks like the lines have a length of something like 1+cosθ, where θ is the angle between the line and the downward vertical. Something to do with polar co-ordinates, anyway.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Your guess is really a nice one!! But it's not how I did it!!

If it is equivalent, this would surprise me.

I'll work out the lengths of the lines and see if it is close to 1 + cos of theta and get back to you.

My unreveiled method uses some other lengths, but not the length of the lines.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Okay, I plotted out the (1 + sin(theta from horizontal)) guess.

And the picture is now not upsidedown for fun like original one.

The bold line is my original MYSTERY CURVE.

The new shell shaped curve with radial lines is the first guess with line lengths of 1 + sin(theta from horizontal).

Should be same as cosine of the phrase "90 minus theta".

Here's the plot below.

*Last edited by John E. Franklin (2006-12-06 03:41:41)*

**igloo** **myrtilles** **fourmis**

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**Patrick****Real Member**- Registered: 2006-02-24
- Posts: 1,005

I have no idea what curve it is, but perhaps the filename can give someone a clue?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Use a taylor series to see if it converges to any known finite polynomial or natural function.

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

http://www.mathsisfun.com/forum/postgallery.php?pid=51418&filename=**SinPlusOneForRadiusVersusAngleProportionalHeight**.png

That might help...

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Oh I didn't know you could read the original filename on the forum!!

That's what I called it because I am comparing (versus) mathysperson's

first guess to my curve.

I think the filename should give away the biggest clue to how I made the curve.

Good Luck with the puzzle...

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Time's Up!!!

This is how it works.

Use degrees, radians, gradians, it shouldn't matter, though I used degrees when I made the diagram.

For the 90 degree angle, straight upward, I drew a line from the center up 9 units in the CAD program.

For the 80 degree angle, I drew a line that extended at 80 degrees from horizontal, but stopped when the end of the line was at the height of 8 units above the horizontal axis.

I continued in this simple fashion.

So for 45 degree angle, the line segment ends where it is 4.5 units above the horizontal.

Pretty cool function, don't you think.

I think I worked it out into polar coordinates, and it had a cosecant in it, but I don't have my notebook out right now... Bye...

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

But then what happens with the 0 degree angle? By continuing the pattern, that should stop immediately (or if it was made to stop at some point above the horizontal, it would go on for ever), but it clearly has a finite length. What calculation did you use there?

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

For zero, I didn't actually do that one, but I did numbers really close like 1 degree and probably one-tenth of a degree on my calculator. Can't remember right now. Sorry. I think it is undefined at zero degrees, but the curve might point right at a number. Maybe we can figure it out. Talk to you later.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

For degrees, it points to 5.729477951 my calculator says from .00001 / sin .00001.

This is the radius at .00001 degrees from horizontal. If I did it right.

For gradians, the curve is enlarged by 10/9, so it hits 6.366197723 or so and (0,10) on the vertical.

I guess that makes sense.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

For the fun of it, I just did it in radians and you won't believe it!!

Check it out for yourself before I give it away...

(smurk)

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Well, here's the answer, time's up again!!

For radians, if I don't divide by ten like the other ones.

Then the vertical distance is pi over 2 and horizontal distance is ONE!!!!

Cool huh!!!l Neato wild, huh???

All these curves are simply close-ups or enlargements of the other ones if you switch between

radians, degrees, and gradians.

Does anyone have any questions about this puzzle??

**igloo** **myrtilles** **fourmis**

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