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One for all of you maths wizzers out there.

I have the length of a chord across a circle and the length of the same

chord around the circle.

How can I find the radius of the circle using Excel?

Try downloading the following file:

cid-f30e41eb2c49c429.skydrive.live.com/.../Excel%20Public/Handy|_07\

|_17|_08.xls

If you are having trouble with the previous link, try:

cid-f30e41eb2c49c429.skydrive.live.com/.../Excel%20Public

then select Handy.

I don't understand "chord around the circle". In my limited

recollection, a chord can only exist inside a circle. a.k.a. a

straight line segment with end points on the circumference.

Could it be the length of one of the (two possible) sections of the

'circumference' which is defined by the chord endpoints?

Unless you're talking music, guitar chords and the circle of fifths.

(which, though I should, I don't know anything about anyway)

You got it right 1st time Steve... I mean the length of the bit of the

cirvumference.

How late? I know I received this at 06:37 but I don't know which bit of the

globe you're on.

I think the OP meant "circumferance" by "chord around circle".

Yes I think we might have a small vocabulary problem here.

Maybe you could check first:

http://www.1728.com/circsect.htm

and then reformulate your question.

Okay guys... Sorry for not being too clear. Not read the whole thing yet but

... At the site Louise gave... I have the distance AB straight across and

the distance ADB... That is the shorter distance around the circumference

from A to B... And what I need is the radius of the circle.

Will go and look closer at the site now... Thanks

The last option is exactly what I want to do!!

Now all I have to do is find out how they do it! :-(

So... Still loking ... Any ideas?

I've given up trying to construct a formula, but I think that if

you're in a hurry the following should do. You'll need to construct a

look-up table of:

theta vs [sin(theta/2) / theta] (theta in radians)

Calculate 1/2 x c/a, find the value of theta from your look-up table

having that value, then calculate r as a/theta

(I'm naming your circle arc "a", chord "c", and "theta" as the angle

subtended by the chord)

Here's the math:

circle arc: a=r x theta (radians)

or r=a/theta

triangle: sin(A)=opposite/hypotenus

or in this case

sin(theta/2)=c/2 / r

doing the math I get: sin(1/theta)/theta = 1/2 x c/a

The following code is what I've gleaned from JS source of the web site

Louise gave me.

I have no idea why the algorythm on that site works. Anyway... My VB

translation don't work :-(

I tested the site calculator by "asking" for the circumference of a circle

of radius 10.

This gave me 62.832.

I halved that and asked for the radius of a circle with a chord length of 20

and a "bow" .. Thanks John .. Of 21.416.

Thee site calculator... Wonder of wonders... Gave me a radius of exactly 10.

My VB code gives 119... Any idea what's going on at all?

Original relevant bit of JS...

--------------------------------

if (chc==0){chord=inp1;arclen=inp2;

arc2crd=inp2/inp1;

centang=4; <!-- radians -->

incr=4;

ct=0;

while (ct<75){

angratio=(centang)/(2*Math.sin(centang/2));

if

(angratio>arc2crd){incr=incr/2;centang=centang-incr}else{incr=incr*2;centang

=centang+incr};

ct=ct+1;}

ct=0;

radius=(chord/2)/(Math.sin(centang/2));

ans[3]=radius;

apo=radius*radius - (chord/2)*(chord/2);apo=Math.sqrt(apo);

ans[4]=apo;

centang=centang*(180/pival); ans[5]=centang;

seghgt=radius-apo;ans[6]=seghgt;

}

----------------------------------

My VBA code

----------------------------------

Sub subCalculateRadius_2()

Const pival = 3.14159265358979

Dim ct As Long

Dim radius As Long

Dim chord As Long

Dim apo As Long

Dim centang As Long

Dim arclen As Long

Dim inp1 As Long

Dim inp2 As Long

Dim chc As Long

Dim sigfig As Integer

Dim incr As Long '<!-- Increment -->

Dim angratio As Long '<!-- Angle Ratio for choice 8 -->

Dim arc2crd As Long '<!-- arc to chord ratio - for choice 8 -->

sigfig = 5

inp1 = 20

inp2 = 31.416

If chc = 0 Then

chord = inp1

arclen = inp2

arc2crd = inp2 / inp1

centang = 4 '<!-- radians -->

incr = 4

ct = 0

Do While ct < 75

angratio = (centang) / (2 * Sin(centang / 2))

If angratio > arc2crd Then

incr = incr / 2

centang = centang - incr

Else

incr = incr * 2

centang = centang + incr

End If

ct = ct + 1

Loop

ct = 0

radius = (chord / 2) / (Sin(centang / 2))

apo = radius * radius - (chord / 2) * (chord / 2)

centang = centang * (180 / pival)

MsgBox "Radius = " & radius & vbCrLf & "Central Angle = " & centang

End If

'********************************************************************

End Sub

That should be 31.461.....................

And guess what... I messed up the nomenclature again.

I mistakenly refered to the "bow" ... It should be the curve ADB.

Can I help it if "bow" implies a curve to me!!!

Anyways... I think by now everyone gets the gist. ... I hope...

Interestingly... If you do choose option Chord & Arc and enter 20 and 31.46

respectively then the apotherm is the same as the triangle area....

0.000036732. where... IMHO they should both be zero.

I think I reacted too promptly, without reading the next posts...

You've done it. And as for the code, I'll have to leave it to other

people.

Too bad for my hope to help...

Another time maybe.

Thank you so much for your help so far!!! Have you anything "to do" with

mathematics? That site was so useful!!

What I would really like to do now but I have no idea how... Is to step

through the JS code at the same time as the VB and check for differences.

I have a minimum of JS from about 8 years ago... I think... Can't

remember... It's not recent anyway... I was trying to help make a friends

web site multilingual... Didn't work because I discovered... That you

need(ed) to write seperate pages for each language.

Ah well.

Again... Your input has been really important.. I hope the JS *and* VBA

people here can sort out what's going on.

I will slog away and post a positive or negative result in any case.

I love anything that has to do with number, and in general, problems

solving.

Was always the first in my math classes.

You would not believe that I even love to prepare my Income Tax

report :-)

My husband found me "maniac" when, in the time where there was no

PCs, I held the family budget to the cent!

But then he found that pretty useful.

But from all the maths, trigonometry was not my preferred. I could do

it, but I am absolutely not "visual", and I preferred algebra much

more.

But it is all so long ago..

My guess is that when conditional expressions are equal, VB defaults

to the Else condition, and JS to the If condition. If you change to

If angratio >= arc2crd Then

then you get the 'correct' answer (though not really)

Your input yields 2=2, thus shunting to the Else condition, which

happens to yield the wrong answer in this case.

Personally, I don't like the algorithm in that it isn't convergent to

the answer, just the increment.

Look at the source.

They did not hide their javascript code.

So I think that, with your programmer experience, (maybe print it

first) you'll be able find the formula your looking for and then

figure out how to write it for Excel.

Since javascript is much more familiar to me then VBA, if you don't

find, tell me exactly which option(s) you would select, and I'll find

the formula for you. I'd be so happy to be able to help for once!

The line ED, referred to as the Sagitta (Latin for arrow) is also called the

Bow of curve ADB.

Yes indeedy ...

You've declared all the variables as Long. This is integer, so you've lost

all the floating point parts.

"ct" and "sigfig" should be declared as Integer. All the rest should be

declared as Double.

Having done that, I get 10.

Note also, that the original has the statement "apo=Math.sqrt(apo);" in it,

and you didn't transcribe that to VBA. It doesn't appear to be used in

calculation of the two results you're using, but it does affect the JS's

ans[6].

> My guess is that when conditional expressions are equal, VB defaults

> to the Else condition, and JS to the If condition.

Not sure I'm understanding you correctly here, but > means "greater than and

not equal" in both languages.

On actually thinking rather than just typing, my comment most

certainly is absurd (A lot of embarrassment here). I noticed that if

the condition >= were used rather than just >, the correct answer was

(almost) obtained. I read your post on variable typing. I rarely

declare variables, (more embarrassment) so I incorrectly assumed that

Long was the same as Double, at least in that the fractional parts

were not discarded. I should've realized this just based on the output

(I'm beet red by now). The incorrect typing would also explain why the

algorithm was not convergent (and sensitive to initial conditions).

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