Equation for Concentric Circles?

Question

I'm trying to work out an equation for the combined value of circumferences of a set of concentric circles.
For example - if we start with a circle of say, 5mm Dia, the Circumfence would be 15.71. If there was then
a circle placed outside of this at a 4mm distance it would have a dia of 13mm - the circumference of this
would be 40.83. Another one at 4mm offset from that would be 21mm (Circumference 65.96). Clear so far?

Well, if you add the above three figures, you get a value (15.71+40.83+65.96) = 122.5. What I want to get is
a simple equation where I can start with a Circle of 5mm, with 2 more Concentric Circles at an offset of 4mm
each time and arrive at the 122.5 I'm looking for. I would want this equation to work for any size circle, with
any number of concentrics and a delta offset (obviously a constant delta, though!). Hopefully I haven't lost you
now...

So for example if you had a circle of 25mm with 15 Concentric circles at an offset of 4mm
(so 25mm, 33mm, 41mm, 49mm) etc. - What would the total value of circumferences be for those 15 circles???

I am doing this so I can work out what the outside diameter a reel of webbing will be after a certain amount of
webbing has been wrapped onto a shaft of a particular diameter.

(Although I guess in reality the pure answer would actually be based on a helix.......)

Answer 1

The first circle will have circumference:

C1 = 2πr

The second circle will have circumference:

C2 = 2π(r + Δr) = 2πr + 2πΔr

where Δr is the increase in the radius from one circle to the next.

The third circle will have circumference:

C3 = 2π(r + 2Δr) = 2πr + 4πΔr

Each successive circumference increases linearly over the previous by the same factor, 2πΔr, so it's an arithmetic progression.

To find the sum of n circumferences, we can use the formula for the sum of an arithmetic series: n(a1 + an)/2

n
Σ Cn = n(2πr + 2πr + 2(n - 1)πΔr)/2
k=1

= nπ(2r + (n - 1)Δr)

So for your example, you were increasing the radius by 4mm each time (Δr), you were doing it 3 times (n = 3), and the initial radius was 5/2 = 2.5 (r):

3π(2 x 2.5 + (3 - 1)(4))
3π(5 + 8)
3π(13)
39π
122.5

So for n = 15 and r = 25/2 = 12.5 and Δr = 4:
15π(2 x 12.5 + (15 - 1)(4))
15π(25 + 14(4))
15π(25 + 56)
15π(81)
1215π
3817.03mm

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Hy forgot to divide the diameter (25mm) by 2 first. (I did this too, but caught it.)

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Scarlett Manuka's mistake was in not noticing that the last term should be:

D + 2(n - 1)d

not

D + 2nd

Although I guess this could be a matter of interpretation.

For your question, are there 15 concentric circles including the middle one, or 16 including the middle one?

If there are 15 including the middle one, Scarlett should have gotten, using his formula:

π (14 + 1) (25 + 14 * 4) = 1215 π = 3817.03 mm

Which is what I got.

If there are 16 including the middle one, Scarlett should have gotten:

π (15 + 1) (25 + 15 * 4) = 1360π = 4272.57 mm

(He would have gotten that number if he hadn't accidentally typed 26 instead of 16 for the first number on the left.)

Answer 2

Let D be the initial circle diameter and d the offset, with n increments.

Then the circle diameters are D, D + 2d, ..., D + 2nd.

The circumference of each circle is πD, π(D+2d), ..., π(D+2nd). The total circumference is
π (D + (D + 2d) + ... + (D + 2nd)).

Fortunately there is a formula to add terms of this type. In this case we get a total of π (n+1) (D + nd).
In your first case n = 2, D = 5 and d = 4. So the total is
π (3) (5 + 8) = 39π = 122.52 mm.
In your second case n = 15, D = 25 and d = 4. So the total is
π (26) (25 + 60) = 2210 π = 6942.9 mm.

(I think the previous answerer forgot that an offset of 4mm implies a diameter increase of 8mm.)

Answer 3

Edit:
My first solution included the circummference of the core, which is incorrect:
Starting with a diameter of D and a thickness t, you have L = π(D + 2t) + π(d + 4t) + π(D + 6t) + . . .
L = π∑D + 2t + D + 4t + D + 6t + . . .
L = π∑nD + 2t(1 + 2 + 3 + . . . + n)
L = π[nD + 2t∑(1 + 2 + 3 + . . . + n)]
L = π[nD + 2t*n(n + 1)/2]

L = π[nD + t*n(n + 1)]

where L = total length
n = number of full wraps
D = the core or starting diameter
t = thickness of the material

To get to the outside diameter, knowing L,
(now it gets fun)
O.D. = D + 2nt

L = π[nD + t*n(n + 1)]
L/π = nD + tn^2 + tn
tn^2 + (D + t)n - L/π = 0
n = (-(D + t) ± √((D + t)^2 +4Lt/π)/(2t)

O.D. = D + 2t(-(D + t) ± √((D + t)^2 + 4Lt/π)/(2t)
O.D. = D - (D + t) ± √((D + t)^2 + 4Lt/π)

O.D. = - t + √((D + t)^2 + 4Lt/π)

Trying these for 15 circles with D = 25 mm and t = 4 mm:
L = 15π[25 + 4*(15 + 1)]
L = 15π[25 + 4*(15 + 1)]
L = 4,194 mm = 4.194 m

O.D. = 25 + 2*15*4 = 145 mm

O.D. = - 4 + √((25 + 4)^2 + 4(4194)(4)/π)
O.D. = - 4 + √(29^2 + 21,360)
O.D. = - 4 + √(841 + 21,360)
O.D. = - 4 + √22201
O.D. = - 4 + 149
O.D. = 145 mm

so the formula, nasty as it looks, really works.

Answer 4

As you say, helix, but concentric circles should give a near enough result.

The radii are in what is called arithmetic progression. Circumference is 2*pi*radius, and if you add up a lot of circumferences you get the same result as if you do

2*pi*(sum of radii)

For n circles, the formula is
Sum of radii = n*(2*first radius + (n-1)*incremental step),

then multiply this by 2*pi to get sum of circumferences.

So, in the 15 circle example you gave,
sum of radii = (15/2)*(2*25 + 14*4) which I think comes to
795.

Multiply by 2*pi, and the total of the circumferences is
4995.132

Answer 5

Let
n = total number of circles
d_1 = diameter of 1st circle (smallest circle)
d_n = diameter of nth circle
f = difference between concecutive circles.

Thus, the total circumference is
C = C_1 + C_2 + C_3 + ... + C_n

Since C_n = π d_n, then
C = π d_1 + π d_2 + π d_3 + ... + π d_n

Thus, we can factor out π
C = π(d_1 + d_2 + d_3 + ... + d_n)

We know that the diameters form an arithmetic progression, so the sum of the n diameters is given by the formula
(d_1 + d_2 + d_3 + ... + d_n) = n[2d_1 + (n - 1)f]/2

Thus,
C = π(n[2d_1 + (n - 1)f]/2)
____________________________
Therefore, the formula for the total circumference is given by
C = nπ/2 [2d_1 + (n - 1)f]

Where
n = number of circles
d_1 = diameter of the first (smallest circle) and
f = common difference (or offset)

^_^