Take a cone of height H and diameter D.

Now spiral a wire about this N times, evenly spaced.

What is the length of this wire?

My approximation is probably close enough, but I wonder if there is a simple answer to this.

(Note: I am NOT planning how to light my Christmas Tree. No, not me.)

 

what is the diameter of the wire? What is your approximate result?

 

What constitutes "evenly spaced"? I can think of at least three reasonable interpretations and I don't think they are equivalent.

Perhaps the simplest definition would be to define the path parametrically in cylindrical coordinates. Imagine a screw that has a pitch such that it goes lengthwise a distance H in N turns. The radius moves linearly with the height. Now you have a path and can integrate the path length over it.

 

Take a cone of height H and diameter D.

Now spiral a wire about this N times, evenly spaced.

What is the length of this wire?

My approximation is probably close enough, but I wonder if there is a simple answer to this.

(Note: I am NOT planning how to light my Christmas Tree. No, not me.)

I'm not sure you can get the integral in closed form, but you can certainly do a numerical integration.

 

πD(N+1)

 

πD(N+1)

How to you figure that?

It would seem to fail two pretty basic limiting cases.

If it were a wire wrapped N times around a cylinder of diameter D (not spiraled, just wrapped) it would be πDN long. So now we make it a very short cone such that it limits to a spiral on a piece of paper and your formula requires that it would require MORE wire to spiral in toward the center.

The other limiting case is that your formula says that the height of the code doesn't matter. But make the height of the cone very tall. The wire has to be at least H long since it goes from the bottom edge to the top point and we can make H larger than your formula.

 

I'd say a fairly decent approximation would be:



This assumes the wire has negligible diameter compared to other dimensions. In other words, the wire is assumed to be infinitely thin.

 

Hi,

If the diameter of the wire is not given i would assume negligible thickness too.

Also, if the wire is closely spaced we can approximate each turn as a circle and then we just need the average diameter to compute the length. The average diameter is 1/2 of the base diameter. If there are N turns and the base diameter is D, then we get pi*D/2.

If the wire is not closely spaced then we have a three dimensional spiral which is a helix, in this case conical.
We know from electrical work that we have solutions that are conical spirals but spaced by an exponential modulator and we integrate that to find other ways of looking at it. so here it will be even simpler because we would just have a linear modulator like A*t or something like that.
We then just use the formula for arc length from calculus to find the total length.

I think this might provide a better and more complete description:
http://mathworld.wolfram.com/ConicalSpiral.html

Notice the similarities between electrical network solutions and the linear conical spiral.

BTW i assume that the turns are evenly spaced.

 

I'd say a fairly decent approximation would be:

View attachment 140646

This assumes the wire has negligible diameter compared to other dimensions. In other words, the wire is assumed to be infinitely thin.

I don't think that limits too well, either.

Consider the case in which H >> D. That expression will limit to H/2, which can't be correct.

 

I'd say a fairly decent approximation would be:

View attachment 140646

This assumes the wire has negligible diameter compared to other dimensions. In other words, the wire is assumed to be infinitely thin.

Hi,

I forgot to comment on that one.

That is a straight line approximation, where the straight line is a ramp that goes from one turn to the next. I get a slightly different result though.
That means if we took one turn and straightened it out we would get a triangle, and the height is H/N and the base is pi*D for the first turn assuming it's the longest one on the bottom of the cone.
Since the average diameter is D/2, the average base length is pi*D/2. Since the height is always H/N, we have as length:
Lavg=sqrt((H/N)^2+(pi*D/2)^2)

Since this simplifies to:
Lavg=sqrt(pi^2*D^2*N^2+4*H^2)/(2*N)

the total length would be:
L=Lavg*N

which gives us the approximation:
L=(1/2)*sqrt(pi^2*D^2*N^2+4*H^2)


We could do a more careful calculation to see how accurate this is.
What might be missing is a small log term, but i have not actually checked this yet.

LATER:
Here is a more complete formula (2nd one below),
This came from defining the space curve in parametric form, then using the formula for arc length. It may simplify i did not check for that, and it may simplify a lot.
Others may wish to double check the second formula.

 

πDN/2

 

πDN/2

What is the basis for claiming that the height of the cone has no impact?

Does it make any sense to claim that you can make a 1 turn spiral up a 10 foot high cone that is 1 inch in diameter at the base and use less than 2 inches of wire?

 

πDN/2

Hi,

Well to elaborate on WBahn's post, if we have a very very thin diameter and 1 turn and a height of 100 inches, the length of the wire is approximately 100 inches. It helps to do a sanity check on these formulas to check them a little.

The average diameter is D/2, and so the average circular loop length around the body of the cone is pi*D/2. Given N turns, we get N*pi*D/2.
We dont need the height there because the height is implicitly included in the average diameter calculation (D/2).

What that formula leaves out though is the extra length it takes to 'climb' up the cone, which adds to the average diameter by some amount. The approximation there is to reckon the actual diameter to be the longest side of a triangle that has a base as long as the average diameter and height the average height, which is simply H/N. We can then use the calculation for the hypotenuse of a right triangle to find the approximate length of the wire for one turn using this second approximate method. It's still approximate though, and these formulas usually end up with some log term in there as the more exact formula has (convert asinh to the log form although the asinh form seems simpler for this problem).

 

I don't think that limits too well, either.

Consider the case in which H >> D. That expression will limit to H/2, which can't be correct.

That's why I called it an approximation. I intended it to be used for the Christmas tree situation.

For a cone 10 units high, 5 units wide at the base and with 3 turns around the cone, here are the results for the exact computation and for my approximation:



Not too bad really, for such a simple approximation. It should be obvious what I did. I just did the calculation for a cylinder and divided by 2. This is better as N gets large.

What MrAl did is the next better simple method, and of course the ultimate is the integration of the arc length formula.

So, ErnieM, what was your approximation?

 

πDN/2 + H/N

H = 10
D = 5
N = 3

L = 26.89

 

That's why I called it an approximation. I intended it to be used for the Christmas tree situation.

For a cone 10 units high, 5 units wide at the base and with 3 turns around the cone, here are the results for the exact computation and for my approximation:

View attachment 140728

Not too bad really, for such a simple approximation. It should be obvious what I did. I just did the calculation for a cylinder and divided by 2. This is better as N gets large.

What MrAl did is the next better simple method, and of course the ultimate is the integration of the arc length formula.

So, ErnieM, what was your approximation?

Hi,

I tried to follow your reasoning but am not sure exactly what you are doing. I guess you used a cylinder?

The 'triangle' hypotenuse solution was:
sqrt(pi^2*D^2*N^2+4*H^2)/2

and the cylinder solution was:
sqrt(pi^2*D^2*N^2+H^2)/2

and notice that the only difference is the "4" in front of the H as a multiplier, which makes the triangle solution always larger, but more important is that the triangle solution always only lags the exact solution by a small amount, while the cylinder solution gets worse for large H. So i was wondering what your reasoning was exactly so i could follow it. I guess you used a cylinder?

Some examples [exact, triangle, cylinder]:
H=10, D=5, N=3: [26.6271,25.5961,24.0866]
H=100,D=5, N=3: [103.6168,102.7383,55.2735]

Note as H gets larger the cylinder solution gets a bit too far off.

As a side note, it is probably easy to add the wire diameter into the mix because that could be said to just increase the diameter.

 

My approximation was to assume the "circumference" of each turn of the spiral was the average of the upper and lower circles. Then add in the height of the turn too.

Tree is 48 inches tall, diameter is 20, and my lights are 12 feet long. Tossing this into my approximation got me to about 4 turns a foot apart.

That gave me a "good nuff" start to play around with.



Me Al and The Electrician probably nailed this, but I would never type that equation into anything to evaluate it without making several mistakes.

Thanks to all for the assist!

(Edited to credit Me Al)

 

The Electrician probably nailed this, but I would never type that equation into anything to evaluate it without making several mistakes.

Thanks to all for the assist!

Actually, MrAl was the one who went to the trouble to look up the details. I just typed it in and evaluated it.

 

Darn. Why couldn't they have these sorts of "real-world" problems in school.

 

My approximation was to assume the "circumference" of each turn of the spiral was the average of the upper and lower circles. Then add in the height of the turn too.

Tree is 48 inches tall, diameter is 20, and my lights are 12 feet long. Tossing this into my approximation got me to about 4 turns a foot apart.

That gave me a "good nuff" start to play around with.

View attachment 140913

The Electrician probably nailed this, but I would never type that equation into anything to evaluate it without making several mistakes.

Thanks to all for the assist!

I think you can have two simple solutions that bound the answer on both sides.

The upper bound is what you have. A lower bound would just be to use the circumference of the midpoint of the upper and lower circles. You best estimate is then just the average of the two. If the error bound is tolerable, then you are done. If not, come up with tighter bounds.