How low does a necklace hang for a given length of chain?

[swanstuff]

This isn't a homework question - just an idle curiosity that sprung up when buying a chain for a necklace for a friend. I thought it would be easy to answer, but I have repeatedly got stuck on it, so if you can help, I would be most grateful. The necklace is now purchased (I guessed!) but the curiosity remains.

The question began as - if I buy a chain of length l, how far vertically below the throat will the top of the pendant be?

I have idealised the situation by assuming that:
● Everything is reduced to two dimensions;
● My friend's neck presents an arc of a perfect circle for the upper part of the pendant to rest on;
● The 'free' part of the pendant forms the two equal sides of an isosceles triangle.
Even with these assumptions, I have not made much headway.

I've done a diagram to illustrate the problem - please see attached and excuse my amateurish sketch (it's my first go on here!)



Points are shown in black. Construction lines and lengths are shown in blue, and angles in red.
In the diagram,

I have shown the idealised circular neck (with centre at O and radius r) in pink, and the necklace chain in black, with the intention being that they coincide along the arc PQR. The pendant hangs at point Z, where each strand of necklace forms an angle γ with the vertical.

I have defined the length of the chain as l = circular arc length PQR + line segment lengths RZ + ZP
Since it is symmetrical, RZ = ZP = PZ so l = PQR + 2 PZ.

N denotes the throat (lowest point of the neck), and it's the distance x = NZ that I really want to calculate in terms of l (although I would settle for d = OZ, as something tells me that may be easier and after all, x = d - r.

I've done all sorts of messing around, and got some fairly easy intermediate results, such as:

x and d are undefined if l < 2πr (and my friend gets strangled), and are zero and r respectively if l = 2πr.
Angle β = 2π - 2α, where α is half of the angle subtended at O by the chord PR.
Circumference of the neck = 2πr, so circular arc length PQR = (β / 2π ) × 2πr = βr.
ZP and ZR are tangents to the circle, and so the angles ZPO and ZRO are right-angles.
The right-triangle ZOP has sides r, x+r=d, and PZ, and so by Pythagoras' theorem, PZ² = r² + (x+r)² .
Therefore l = βr + 2 sqrt( r² + (x+r)² ).
And so on ... I found formulæ for sin α in terms of PZ and x+r, found that γ = ½ (β - π),
I tried expressing P in terms of polar coordinates, and working out the gradient of PZ, and plugged all sorts of things into other things.
I felt that the key was to find simultaneous equations for PQ - one based on the circle above this chord, and one based on the triangle below, but was unable to make any headway after doing so.
I had a sinking feeling that calculus may come into it somewhere ... something like "what is the maximum of x as α varies from 0 to 2π ... but I could not see how that would force ZP and ZR to be tangents, and I don't want the necklace cutting into my friend's neck!

Anyway, after 15 sheets of paper, I decided that it was beyond me, so if anyone can help, please let me know what is the formula for x, if you are given values for l and r?

Thank you!

 

[Cubist]

That's a great diagram! But the bad news is that I'm pretty sure there's no algebraic solution for this problem. I did...

l=r*β + 2*PZ
l = r*(2*π - 2*α) + 2*PZ
l = 2*r*(π - α) + 2*r*tan(α)
l = 2*r*(π - α + tan(α))
l/(2*r) - π = tan(α) - α

...when I saw the RHS I stopped working. Seeing something of the form tan(x)±x makes me think that this can only be solved numerically. You can use a spreadsheet's solver, Newton Raphson, binary chop method, etc to find an arbitrarily close approximation to the above, obviously in the range 0≤α<π/2. Once you've found a value for α then

x = r*( sec(α) - 1 )

--

If you prefer to eliminate α...

l/(2*r) - π = tan(α) - α
l/(2*r) - π = sqrt( (r+x)² - r² ) - arccos(r / (r+x))

tan(α) was obtained using Pythagoras in triangle OPZ, and α obtained from cos(α) = r / (r+x).

Finding a value for x still requires a numerical solver, with values of x≥0.

 

[swanstuff]

Hi Cubist. Thank you for the kind words and for working on this problem.

I was getting similar reductions, with the angle appearing both 'naked' and inside a trig function. It's been so long since I did any trig that I thought it was just a formula that I'd forgotten and couldn't find that was preventing me from getting it down to just one α and progressing to a formula for x.

Your reply, while disappointing that it didn't have a nice neat answer for me, was at least encouraging that I hadn't simply blundered or been ignorant, but had in fact got stuck somewhere that stumped others too.

For such action(relatively) simple question, though, it seems strange that there's no 'easy' answer ... Unless someone else knows better.

Thank you - Steve

 

[Dr.Peterson]

For such a (relatively) simple question, though, it seems strange that there's no 'easy' answer ... Unless someone else knows better.

It's easy to ask a question that's hard to answer -- just listen to any 3-year-old!

Teachers tend to give students the impression that math can do anything, by showing only the problems that can be solved by the methods they teach. They don't tend to mention that most polynomials you can write can't be factored, and most calculations require rounding, and most equations you can write (or that arise in the real world) can only be solve by approximate numerical methods! Simple formulas are actually rare.

 

[swanstuff]

Hi again - and thank you Dr Peterson: that makes sense, I suppose - teachers will generally only set problems that their students have a chance to solve (at least in the less rarified levels of study) so we all get used to expecting to find neat answers!

I thought some more about this problem, based on Cubist's suggestion of a numerical method. As my Excel doesn't seem to like add-ins, I turned it round to what seemed easier - setting the required x (the depth of the pendant below throat level), and then work back to find what value of L (total length of necklace: I've changed my symbol from l so as to avoid confusion with 1). This would require setting a value for r (neck radius), but I decided it would be better to instead fix a value for c, the effective neck circumference, as it is easier and far less painful to measure this on a real human neck. I also decided to do the calculation in a special non-SI unit that I’ve called sexagenajows*, or sj for short. Very conveniently, my friend’s neck has an effective neck circumference of 1 sj.

c = 1 sj

c = 2πr

so r = c/2π = approx. 0.159 sj

Referring to the original diagram, and remembering that d = r + x:

by Pythagoras on triangle ZOR, d² = r² + {½[L-rβ]}² ……………………… [1]

β = 2π - 2α = 2 (π-α)

also, cos α = r/d, so α = arccos (r/d)

therefore β = 2 (π - arccos (r/d))

Substituting this back into [1] gives

d² = r² + { ½ [ L - [2r ( π - arccos (r/d) ) ] ] }² ……………………… [2]

Rearranging:

d² - r² = { ½ [ L - [2r ( π - arccos (r/d) ) ] ] }²

sqrt (d² - r²) = ½ [ L - [2r ( π - arccos (r/d) ) ] ]

2 sqrt (d² - r²) = L - [2r ( π - arccos (r/d) ) ]

L = [2r ( π - arccos (r/d) ) ] + 2 sqrt (d² - r²) ……………………… [3]

Using r = c/2π and d = r+x so that d = x + c/2π back into [3] gives

L = [2(c/2π) ( π - arccos ((c/2π)/( x + c/2π)) ) ] + 2 sqrt ((x + c/2π)² - (c/2π)²) ……………… [4]

This can be tidied up a bit, by setting c = 1 sj as above, and replacing the constant c/2π = 1/2π by the symbol w, to give L as a not-too-intimidating function of x:

L = [2w ( π - arccos (w/( x + w)) ) ] + 2 sqrt ((x + w)² - w²) ……………… [5]

Let’s plug in some “x”s and see what Ls come out!

* This is a made-up unit. A “jow” is an Indian unit of length, now obsolete, equal to about ¼ inch or 0.63 centimetres. “Sexagen-” is a prefix denoting a multiple of sixty (cf. sexagesimal), and I estimate 60 jows (15 inches or 38 cm) to be a not unreasonable ‘effective circumference’ of the human neck, remembering that even a tight necklace will naturally lay higher at the nape than the throat. The idea is that if anyone actually wants to use this in practice, all they have to do is to measure the effective circumference of their friend’s neck in whatever units they prefer, and then remember that this number is equal to 1 sexagenajow in their units. Suppose it comes to 30 cm, then 32cm = 1 sj. If they want a pendant that hangs 16cm below their friend’s throat, (ie ½ sj) then they can look in the table and see that if x = 0.5 sj, then L =1.86 sj (approx), so they will need a chain that is 1.86 × 32 = 59.4 cm long.

 

[swanstuff]

First a small correction to my earlier post: It should have said:
Suppose it comes to 30 32cm, then 32cm = 1 sj.

This table gives values for x (the 'drop') and then the required necklace length (L) which will produce that drop. The last two columns show the two components that make up L, the first (PQR) being the loop around the back / sides of the neck, and the second being the free-hanging part on the chest (PZR). Values of x above about 1 are not practical, but are shown for interest. Any x longer than about 3½ is liable to cause a trip hazard as the pendant will dangle along the ground when walking. Above about 5 will be a trip hazard to other pedestrians, and necklaces with a very excessive x and L will be less of an item of jewellery and more of a ball-and-chain due to their weight!

____x_______L______PQR___PZR___

0.0000 1.0000 1.0000 0.0000
0.0250 1.0175 0.8322 0.1853
0.0500 1.0467 0.7753 0.2714
0.0750 1.0814 0.7379 0.3435
0.1000 1.1195 0.7105 0.4091
0.1250 1.1600 0.6892 0.4708
0.1500 1.2022 0.6721 0.5301
0.1750 1.2457 0.6580 0.5876
0.2000 1.2901 0.6461 0.6439
0.2250 1.3352 0.6360 0.6993
0.2500 1.3810 0.6272 0.7539
0.2750 1.4273 0.6195 0.8079
0.3000 1.4741 0.6127 0.8614
0.3250 1.5211 0.6066 0.9145
0.3500 1.5685 0.6012 0.9673
0.3750 1.6161 0.5963 1.0198
0.4000 1.6639 0.5919 1.0721
0.4250 1.7119 0.5878 1.1241
0.4500 1.7601 0.5841 1.1760
0.4750 1.8085 0.5808 1.2277
0.5000 1.8569 0.5776 1.2793
0.5250 1.9055 0.5747 1.3308
0.5500 1.9542 0.5721 1.3821
0.5750 2.0029 0.5696 1.4334
0.6000 2.0518 0.5672 1.4846
0.6250 2.1007 0.5651 1.5357
0.6500 2.1497 0.5630 1.5867
0.6750 2.1988 0.5611 1.6377
0.7000 2.2479 0.5593 1.6886
0.7250 2.2970 0.5576 1.7394
0.7500 2.3462 0.5560 1.7902
0.7750 2.3955 0.5545 1.8410
0.8000 2.4448 0.5531 1.8917
0.8250 2.4941 0.5517 1.9424
0.8500 2.5435 0.5504 1.9931
0.8750 2.5929 0.5492 2.0437
0.9000 2.6423 0.5480 2.0943
0.9250 2.6917 0.5469 2.1448
0.9500 2.7412 0.5458 2.1954
0.9750 2.7907 0.5448 2.2459
1.0000 2.8402 0.5438 2.2964
2.0000 4.8300 0.5235 4.3066
3.0000 6.8263 0.5160 6.3103
4.0000 8.8244 0.5122 8.3122
5.0000 10.8232 0.5098 10.3134
6.0000 12.8224 0.5082 12.3142
7.0000 14.8218 0.5071 14.3148
8.0000 16.8214 0.5062 16.3152
9.0000 18.8211 0.5055 18.3155
10.000 20.8208 0.5050 20.3158
20.000 40.8196 0.5025 40.3171
30.000 60.8191 0.5017 60.3175
40.000 80.8189 0.5013 80.3177
50.000 100.8188 0.5010 100.3178
100.00 200.8186 0.5005 200.3181
1000.0 2000.8183 0.5001 2000.3183
10000. 20000.8183 0.5000 20000.3183
100000 200000.8183 0.5000 200000.3183

Exactly as I would have expected, PQR tends towards 0.5 - that is to say, as the pendant hangs lower and lower, the 'free hanging' parts of the necklace become almost parallel, and the tangents they form touch the neck/circle at points almost diametrically opposite, dividing the circle in two.

Also as expected, L (and eventually PZR) tends towards 2x, which is to say the loop around the back of the neck becomes negligible, and an increase of 1 sj in the length of the chain is shared between the two strands on either side, so that the pendant drops 0.5 sj.

 

[Cubist]

While checking your post#5 I spotted a mistake in my post (the red part was missing)...

l/(2*r) - π = tan(α) - α
l/(2*r) - π = sqrt( (r+x)² - r² )/r - arccos(r / (r+x))

Now we get the same result! I need to remember that tan is opposite over adjacent. By the time I'd worked out what the "opposite" part was I'd long forgotten about the "over adjacent" ?