NEED HELP ON MATH PROBLEM

[CAD CHICK]

I need help trying to figure out the radius of a line that travels at an agle around a cylinder. (this is for some spiral stairways, I need to know what roll radius to give our fabricator)

Some one gave me this formula, but I'm not 100% sure it's correct

c=hypotenuse a=width b=height
r=radius of lenght traveled (this is what I need)
VS=is the distance between the midpoint of my arc and the midpoint of "a"

c=sq root of (a^2=b^2)
r= (4(VS^2)+c^2) / 8(VS)

Thanks in advance.

I have a picture of what I need i can e-mial it if you need clarification.

 

[tkhunny]

I'm not real clear on the use of the word "radius". Why would the radius change with a different angle of wrap? The radius of the cyllinder is the radius of the spiral. Do you mean the length of the spiral?

If the spiral just went around the cyllinder, without changing elevation, the length, obviously, is 2*r*pi, where 'r' is the radius of the cyllinder.

Just some starter thoughts to see if we can get on the same page.

 

[galactus]

This appears to be a circular helix. We can use parametrics do solve it. But, like tkh, I would also like to be on the same page.

Excuse my sloppy helix. I did the best I could in Paint.

If we had '?', as I marked in the diagram, and perhaps the length of the spiral, we could maybe find the radius.

Here is an example of what I mean.

"Suppose we wrapped half inch diameter tubing around a cylinder which has diameter 12 inches. What length of tubing will make one complete turn around the cylinder in a distance of 20 inches?".

This can be solved using parametrics and the arc length formula. Is your staircase something like this. I am having a rough time interpreting your description.

 

[CAD CHICK]

it is a helix, and that's where the radius changes it starts at 0 at the bottom of the cylinder then it goes to a 32'-0" elevation wraping around the tank.

 

[tkhunny]

Immediate agreement on the defintion of C2. That appeaars to be correct.

R1 appears to be the "Radius" I've been talking about. The fixed radius of the cylinder around which the helix travels.

I just don't understand R2. If it truly is a radius as it is drawn, it can't be fixed for the entire helix. It starts out at zero (0) and ends at "Chord In Plan".

 

[CAD CHICK]

Thats what I was not sure, but I think the formula is right. A friend of mine told me to use the R1*2= R2 just for fabrication of my stairs and is close to being right. But I wanted to verify the answer with a more accurate one.

 

[tkhunny]

One more quick thought short of a solution.

We know that:

\(\displaystyle (R1-VS)^{2}\;+\;\left(\frac{CIP}{2}\right)^{2}\;=\;R1^{2}\)

This leads to:

\(\displaystyle R1\;=\;\frac{4*VS^{2}+CIP^{2}}{8*VS}\)

This looks sufficiently similar to the solution for R2 that I should be either impressed or suspicious.

However, I have a strong desire to redefine your drawign just a bit so that we can talk about one other aspect.

In a 3-dimensional Coordinate axis system:

Put the bottom point of the helix at (R1-VS,-CIP/2,0)
Put the top point of the helix at (R1-VS,CIP/2,RISE)
Put the central point of the helix at (R1,0,RISE/2)

This should be sufficient to define a plane in which a circle contains the three given points. The new plane is slanted (not parallel to any of the x-, y-, or z-planes we started with). We should also be able to find the center of this circle.

After constructing spheres of equal radius on each of these three points, I have a very strong desire to investigate the point (-19.948052,0,16)

Note: \(\displaystyle -19.948052\;=\;\frac{-RISE^{2}}{8*VS}\)

Oddly enough, with R1 = 18.766473 (calculated as shown above), this gives R1 + 19.948052 = 38.714525

That may not seem significant until you notice this:

\(\displaystyle C2\;=\;\sqrt{CIP^{2}+RISE^{2}}\;=\;42.692519\)

\(\displaystyle R2\;=\;\frac{4*VS^{2}+C2^{2}}{8*VS}\;=\;38.714525\)

Now we should be impressed.

We also should be impressed that the graphical depiction of R2 is not helpful. We are looking for R2, the radius of curvature of the helix. The center of this radius of curvature should not be particularly in the neighborhood of the cylinder.

It is a confusing drawing, but I do believe we have demonstrated the efficacy of the formula you have been given. Go and build a giant stair case!

 

[tkhunny]

It would be irresponsible of me to claim to be supremely confident of this result. In my mind, there remains a uniqueness problem. The equations I used actually produced four solutions. Two were easily discarded because R2 < 0. This left me with two solutions. One was compelling because it matched the results provided in the original problem. I have not thought through why the other solution would not also be acceptable.

In this case, the solution provided above has x-displacement > R1. The other solution has x-displacement < R1. That may be significant.

Maybe it will come to me.

 

[CAD CHICK]

Thank you a bunch to all of you, you don't know how much I appreciate all of your help and knowledge.