Mathfilter - Surface of a Helix

June 28, 2015 4:15 AM

My math is a bit rusty, I would like to calculate the surface of a helix

It looks similar to this flat ribbon

I assume zero thickness.
I want only the surface of one side
I want to adjust the length in the formula
I want to adjust the radius in the formula but the helix is always 1/2 of diameter.
I want to be able to adjust the pitch.
The helix does not have to have an integer number of full turns.

Any ideas?

posted by yoyo_nyc to Science & Nature (3 answers total)

I think your first task is to come up with a way to parameterize your surface. If you use the formulation here and then add a second variable s that you multiply to the radius to denote the location of a point along the "width of the ribbon", that should do it.

On the other hand, if all you want to do is find the surface area of your surface, then I think it will suffice to find the arc length of the corresponding helix (that is, the one-dimensional curve corresponding to, for example, the outer edge of your surface) and then multiply by the width of the ribbon.
posted by J.K. Seazer at 5:06 AM on June 28, 2015

Ah, never mind, not quite as simple as multiplying. The arc length of the helix increases with radius. So, given that your minimum radius from the axis of your helix is m, and the maximum radius is M, you would want to integrate the helix arc length formula T*sqrt(a^2+b^2) for a = m to M.
posted by J.K. Seazer at 5:16 AM on June 28, 2015 [1 favorite]

Best answer: Suppose that p is the how much height the helix gains as it turns through one radian.

Let's imagine a small part of the helicoid: between r and r+δr, and as it turns through an angle δθ. The area of the small piece, which is approximately rectangular, can be calculated as follows:

Its length in the radial direction is just δr.
Its width in the angular direction can be calculated by Pythagoras' theorem: it goes a distance r*δθ across, and p*δθ up, so the width is sqrt(r²(δθ)² + p²(δθ)²).
So the area of this small piece is δr*sqrt(r²(δθ)² + p²(δθ)²).

To find the total area, we just integrate this. So if R is the maximum radius, and the helicoid turns through a total angle of ϴ (that's a capital theta), the surface area is the integral of sqrt(r²+p²) dr dθ as r goes from 0 to R and θ goes form 0 to ϴ. Doing the integral gives the final formula:

Area = ϴ/2 * [R*sqrt(R²+p²) + p² * ln(R/p + sqrt(1+R²/p²)) ].

If you don't like thinking about radians: let h be the height that the helicoid gains on each full turn, and let T be the total number of turns (which may be fractional). p = h/(2π), and ϴ = 2πT, and the area is

Area = T/2 * [R*sqrt(4π²R² + h²) + h²/(2π) * ln(2πR/h + sqrt(1 + 4π²R²/h²)) ].

If you want to search the internet for better explanations and probably some pictures, the term you want is "helicoid" (as opposed to helix).
posted by water under the bridge at 7:04 AM on June 28, 2015 [5 favorites]