/sci/ - Science & Math » Thread #14381696

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Nasty problem

Anonymous Sun 10 Apr 09:12:18 2022 No.14381696 View Reply Original Report

Quoted By: >>14381713 >>14381714 >>14381718 >>14381725 >>14381729 >>14381854 >>14381996 >>14382042 >>14382729 >>14383764 >>14383775 >>14383861 >>14386259 >>14387100 >>14387830

Hi there, I stumbled across this problem at work, if solved it will simplify my work quite heavily.

Is it possible to find the radius of a circle having only the length of 2 parallel chords?

(the problem originally is 3d, find the sphere radius having only the parallel inscribed discs)
Any hint is greatly appreciated

Anonymous

Anonymous Sun 10 Apr 2022 09:26:02 No.14381707 Report

Quoted By: >>14381748

i mean, are you aware of the distance of the parallel chords from the center?

I'm guessing otherwise it would be impossible

Anonymous

Anonymous Sun 10 Apr 2022 09:30:11 No.14381713 Report

Quoted By:

>>14381696
You could obviously fit many different circles into the same, so it can't be possible.

Anonymous

Anonymous Sun 10 Apr 2022 09:30:31 No.14381714 Report

Quoted By:

>>14381696
if we know nothing more than the fact that chords are parallel to each other, then we can't solve it. easy as that.

Anonymous

Anonymous Sun 10 Apr 2022 09:32:18 No.14381718 Report

Quoted By:

>>14381696
The lengths of two parallel chords is not enough data.
Every circle is going to have chords ranging from zero to the diameter of the circle so any pair of numbers can describe chords in any circle with a diameter greater than the largest chord. You can do something like find the minimum such circle, but that is trivially setting the largest chord as the diameter.

Anonymous

Anonymous Sun 10 Apr 2022 09:35:10 No.14381725 Report

Quoted By: >>14381767 >>14387833

>>14381696
To expand upon what other anons said, if we do not know the distance, we have a lower bound for the radius (diameter = the larger cord) but no upper bound.

Anonymous

Anonymous Sun 10 Apr 2022 09:37:37 No.14381729 Report

Quoted By:

>>14381696
0*?=q*r=f*g

Anonymous

Anonymous Sun 10 Apr 2022 09:39:15 No.14381732 Report

Quoted By: >>14381735 >>14381738 >>14381743

If you have two parallel chords, YES you can find the diameter of the circle.
Two line segments give four endpoints; four known points in the plane is enough to define a circle.

Anonymous

Anonymous Sun 10 Apr 2022 09:41:51 No.14381735 Report

Quoted By:

>>14381732
but of course only if you know both the length and position of the chords. Otherwise he only effectively knows three points.

Anonymous

Anonymous Sun 10 Apr 2022 09:42:31 No.14381738 Report

Quoted By: >>14382621

>>14381732
OK, show me. With chords 2cm and 5 cm long, for example.

Anonymous

Anonymous Sun 10 Apr 2022 09:46:59 No.14381743 Report

Quoted By: >>14382450 >>14382625

>>14381732
OP here, yes but only if you have the points coordinates, here I only have the lenghts

Anonymous

Anonymous Sun 10 Apr 2022 09:49:59 No.14381748 Report

Quoted By: >>14381751 >>14381752 >>14381755 >>14381767 >>14381822 >>14383807

>>14381707
OP here, no that's the problem...
However it should be possible since there is only one circle with 2 chords

Anonymous

Anonymous Sun 10 Apr 2022 09:51:04 No.14381751 Report

Quoted By:

>>14381748
edit, with those two chords

Anonymous

View Same Google iqdb SauceNAO Screenshot 2022-04-10 115216.png, 101KiB, 966x519

Anonymous Sun 10 Apr 2022 09:52:29 No.14381752 Report

Quoted By: >>14382074 >>14386212

>>14381748
But it isn't, you can have different diameters with same chords

Anonymous

Anonymous Sun 10 Apr 2022 09:54:18 No.14381755 Report

Quoted By:

>>14381748
It's not possible. Take the longer cord and create any arbitrry circle with it. Now imagine sliding that cord up or down the circle until the length matches up with the shorter one. Now imagine starting with a different arbitrary circle and doing the same. There is no unique solution.

Anonymous

Anonymous Sun 10 Apr 2022 09:59:11 No.14381767 Report

Quoted By: >>14381776

>>14381748
see >>14381725
Do you need proof or is it trivially obvious?

Suppose we have two chords $a,b$ and a diameter $d$ such that $d\ge a\ge b$ .
We can parametrize a circle using sine and cosine and easily see that for each circle of diameter $d$ we can solve
$\frac12 d \cos(\phi) = \frac12 a \Rightarrow \phi = \arccos\left(\frac{a}{d}\right)$ Hence as long as d is large enough, we can find a pair of chords of your length of choice.

Anonymous

Anonymous Sun 10 Apr 2022 10:02:03 No.14381776 Report

Quoted By:

>>14381767
$\frac12 d \cos(\phi) = \frac12 a$
$\Rightarrow \phi = \arccos(\frac{a}{d})$

Anonymous

Anonymous Sun 10 Apr 2022 10:31:42 No.14381822 Report

Quoted By:

>>14381748
how do you not know where the center is? Do you know how far apart the chords are? if so then yes there should be a way to derive a curvature of its parent sphere.

Anonymous

Anonymous Sun 10 Apr 2022 10:53:58 No.14381854 Report

Quoted By:

>>14381696
A circle is homoemorphic to a triangle which means it needs at least three points to be measured, where for the definition of a circle the measurement is on the radius.

Anonymous

Anonymous Sun 10 Apr 2022 10:57:42 No.14381857 Report

Quoted By:

For parallel disks inscribed in a sphere the case is even simpler since you have more than 4 points, actually infinite amount of them.

Anonymous

Anonymous Sun 10 Apr 2022 11:03:49 No.14381868 Report

Quoted By:

You need the distance between them. The bigger the distance - the smaller the circle.

Anonymous

Anonymous Sun 10 Apr 2022 11:05:22 No.14381875 Report

Quoted By:

Interesting observation using continuity and methods from calculus is that a sphere can be viewed as a continuity of circles inscribed in it in such way that two given disks intersect with the circles and therefore define their radius.

So all the circles that intersect with the disks have their surface points defined, which assures even more than the sphere has it's radius defined

Anonymous

View Same Google iqdb SauceNAO construction.png, 38KiB, 892x821

Anonymous Sun 10 Apr 2022 12:02:36 No.14381973 Report

Quoted By:

i can construct a circle from the two segments, don't know if that helps

Anonymous

Anonymous Sun 10 Apr 2022 12:13:00 No.14381996 Report

Quoted By:

>>14381696
You need to know something about the spacing of the chords otherwise a larger difference could mean larger separation began then or a larger radius.

Anonymous

Anonymous Sun 10 Apr 2022 12:38:23 No.14382042 Report

Quoted By:

>>14381696

You need to know the separation between the chords

Your CAD software will have already told you this

Anonymous

Anonymous Sun 10 Apr 2022 13:14:47 No.14382074 Report

Quoted By: >>14382102

>>14381752
What'd you use to make that? It looks pretty

Anonymous

Anonymous Sun 10 Apr 2022 13:29:39 No.14382102 Report

Quoted By:

>>14382074
Fusion 360

Anonymous

Anonymous Sun 10 Apr 2022 15:46:32 No.14382450 Report

Quoted By:

>>14381743
>here I only have the lenghts
Just post the homework problem because its obvious by you're spelling that you're leaving out context.

Anonymous

Anonymous Sun 10 Apr 2022 16:48:03 No.14382621 Report

Quoted By: >>14384860

>>14381738
you didnt specify their relative position in the plane

Anonymous

Anonymous Sun 10 Apr 2022 16:49:03 No.14382625 Report

Quoted By:

>>14381743
build a coordinate system faggot. do you know their positions? thats all you need

Anonymous

Anonymous Sun 10 Apr 2022 17:16:52 No.14382729 Report

Quoted By:

>>14381696
OP, imagine you have in your hands 2 disks of different size. Imagine a hollow half-sphere bowl with a diameter greater than ur two disks. Imagine placing the smaller disk first in the bowl perpendicular to the ground, then laying the larger disk in the bowl.
It should be obvious that they would both fit cleanly in the bowl and there is a space between them. I never said the diameter of the bowl, only that its bigger than the disks. Thus, there are an infinite number of half-spheres that solve ur problem.

On the other hand, do you know the SPACE between the two disks? Knowing the distance between the two would force the sphere to be rigid and have one size. IMO, it'd prob be easier to start with the sphere radius R, and the two disk radii r1,r2 with r1>r2 and solve for distance D and than reverse everything to solve for R.

$\sqrt{R^2-r_2^2} - \sqrt{R^2-r_1^2}=D$
Solving for R can get a couple diff stuff like
$R^2 = (\frac{r_1^2-r_2^2+D^2}{2D})^2+r_2^2$ or
$R^2 = (\frac{r_1^2-r_2^2-D^2}{2D})^2+r_1^2$ or
$R^2 = (\frac{r_1^2-r_2^2}{2D})^2 + \frac{D^2}{4} + \frac{r_1^2+r_2^2}{2}$

Anonymous

Anonymous Sun 10 Apr 2022 22:39:15 No.14383764 Report

Quoted By:

>>14381696
If they're fixed chords take any 3 of the 4 endpoints and construct the circle containing them in the usual way by intersecting perpendicular bisectors of segments between pairs of them.

Anonymous

Anonymous Sun 10 Apr 2022 22:42:44 No.14383775 Report

Quoted By:

>>14381696
uh, the diameter?

Anonymous

Anonymous Sun 10 Apr 2022 22:53:47 No.14383807 Report

Quoted By:

>>14381748
that's not true. do you know how far apart those two chords are?

Anonymous

Anonymous Sun 10 Apr 2022 23:18:16 No.14383861 Report

Quoted By:

>>14381696
>(the problem originally is 3d, find the sphere radius having only the parallel inscribed discs)
i think you can find the answer in books about this subject
>Plane and Spherical Trigonometry by Edward Brooks 1891
Spherical Trigonometry is what you're looking for, they don't teach in schools anymore

Anonymous

Anonymous Mon 11 Apr 2022 07:48:40 No.14384762 Report

Quoted By: >>14386212

Given d (diameter) > a > b, could someone draw me two circles that fit this? I'm having trouble visualizing it.

Anonymous

Anonymous Mon 11 Apr 2022 07:56:54 No.14384774 Report

Quoted By: >>14386212

without changing the length of a or b

Anonymous

Anonymous Mon 11 Apr 2022 08:52:27 No.14384860 Report

Quoted By:

>>14382621
That was the point.

Anonymous

Anonymous Mon 11 Apr 2022 18:51:51 No.14386212 Report

Quoted By: >>14387853

>>14384762
>>14384774

See
>>14381752

Anonymous

Anonymous Mon 11 Apr 2022 19:09:13 No.14386259 Report

Quoted By:

>>14381696
Yeah, there's a way. You need to calculate the circle based on the arcs at the end of the two lines. How you do this, I don't know. I skipped trig. I'll work on it for a little bit and see if I can come up with something.

Anonymous

Anonymous Mon 11 Apr 2022 21:27:18 No.14386702 Report

Quoted By:

If the perpendicular distance from the circle to the midpoint of the smaller chord is the same as the distance from the midpoint of the smaller chord to the midpoint of the larger chord, as appears to be the case in the picture, the problem might be solvable.

Anonymous

Anonymous Tue 12 Apr 2022 00:39:12 No.14387100 Report

Quoted By:

>>14381696
all you can determine based on the conditions is:
2r >= f

Anonymous

Anonymous Tue 12 Apr 2022 08:31:41 No.14387830 Report

Quoted By: >>14387833

>>14381696
I think you can calculate lower and upper bounds on the radius. Probably. But you have an infinite number of circles that can satisfy those requirements.
You'll end up with something like:
a <= r <= b
where a, b in R

Anonymous

Anonymous Tue 12 Apr 2022 08:33:54 No.14387833 Report

Quoted By:

>>14381725
>lower
>>14387830
oh yeah, agree. Lower bound only. Upper bound impossible.

Anonymous

Anonymous Tue 12 Apr 2022 08:55:02 No.14387853 Report

Quoted By: >>14387939

>>14386212
oh I meant to say if a and b were on the same side of the circle
https://www.geogebra.org/m/wat29ggn
I figured it out tho

Anonymous

Anonymous Tue 12 Apr 2022 09:58:48 No.14387939 Report

Quoted By: >>14387945

>>14387853
bro are you blind?
that's exactly what my image shows

Anonymous

Anonymous Tue 12 Apr 2022 10:06:13 No.14387945 Report

Quoted By:

>>14387939
oh i am blind

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