[Closed] bit of maths help.....

I'm not sure i ujnderstand the terms you're using.

i have two circles drawn on a piece of paper showing a solid ring. THe outer diameter (distance from one side to the other in a straight line) is 10.
The inner diameter from one side to the other is either 7 or 8 depending on what you're referring to.
The Chord for a single circle is the same as its diameter no?

Likewise confused. Got a diagram?

I had to look it up.

[i]annulus
?anj?l?s/
nountechnical
a ring-shaped object, structure, or region.
MATHEMATICS
a plane figure consisting of the area between a pair of concentric circles.[/i]

[img] [/img]

[s]I get the question but don't know the answer

In this image the outer chord is 10 - shown by the lin.

Now imagine a line from the end of the chord going to the centre of the circle. where the line crosses the inner circle, we have another (inner) chord. the inner cord is 8

how wide is the coloured circle[/s]

No I don't get the question. however this page mighthelp

http://www.thingiverse.com/thing:377173

Ah ha a response!, basically the outer ring is fixed at 10 and the inner ring is unknown the chord can vary in size from 10 (obviously the center)to effectively zero (the edge of the circle) so knowing the length of the chord tells me where it is on the outer circle. Now if i measure the inner circle on the same chord it will give me the width of the anulus at that point but will be an angular measurement and not the true distance if measured as two concentric circles if that make sense?? am i way off in what is needed to calculate it? ie do i need to know the inner diameter? ta.

How on earth do you get a chord of 10 inside a circle with a circumference of 10?

I don't know the answer, but I would guess you need to use some arc formula if you know the diameter is 10 & chord 8, then you have a difference which gives you another dimension to work with.

Do you want to know the relationship between the chord length and the distance from the chord to the outer circumference?

perhaps a labelled diagram might help? Do you have one to share?

I thought 'not very good with maths' was going to lead to a much simpler question than you asked.

#leavesthread 😳

I'm not sure if I understand the question fully but if you're trying to figure the anulus width you need with inner radius (r). You have the outer radius (R) and the chord length (let's call it 2L).

So if you draw the radius lines from the centre to the ends of the chord you have right angled triangles where the hypotenuse is R and the opposite side r and the adjacent side is L. So its just trigonometry.

Working it through if big circle is diameter 10 then R = 5
Chord length is 8 then L = 4

Then inner circle radius is r = sqrt(R^2-L^2) = sqrt(25-16) = 3

So the anulus width is R-r = 2.

Was that the question? 😀

This obviously assumes the chord is tangent to the inner circle which I'm unsure from the question whether this is the case...

Erm, I should be working, but can you use the properties of an arc (using radians) vs a chord (using trig) and then find the radius of the inner circle (the tangent of the chord).

Of course I may be missing something and I may have misunderstood the Q 😆

The diameter is 10! so the max the chord can be is 10 ie the centre line. from the helpful picture above looking at the middle drawing if you imagine that the outer edge is always the same diameter but the inner varies and the chord also varies so basically i can work out where the chord is on the circle by measurement (say 8 ) so i know its not on the centre line i can also measure the distance from where the chord intersects the outer circle and the inner circle but this is not the true distance as measured form the centre of the circle between the two circles. i hope that makes sense....

there's always a triangle between centre and each of the ends of the chord

2 of the sides of that triangle are fixed at 5 (radius to the outer edge), and the length of the other is your chord

that triangle can be "halved" to make two right angled triangles, hypotenuse of 5 and one side is (chord/2). the other side is the inner radius

5^2 = (chord/2)^2 + radius^2

so inner radius = sq root of [ 25-(chord/2)^2 ]

??

Could you maybe start a thread on help with Grammar first? Then we might be able to understand what you are asking?

Thanks Chilled, I appreciate your help..

I've just re-read and read the whole thread and get the question now, agree with Scardeypants solution.

2 x sqrt(5^2 - r^2)= Chord length

r in the above is the small circles radius.

If I've read the question correctly the answer is 0.3. The chord isn't a tangent to the inner circle so you use the outer circle to determine the distance from the centre to the chord (Pythagoras) the use this distance to calculate the radius of the inner circle. (Pythagoras again).