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'[OT]: Calculating a radius'
2002\04\03@062404 by bbj

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Greetings you lot in [OT] land.

First, i dont get [OT] messages so please reply to me in person.

Now, for the question, one that i could probably solve while i was still at
school, but alas, that was many moons ago.

I have a bow ( u know - like in a bow and arrow ).

The lenght of the piece of "wood" and the length of the piece of string is
known.
Assume that the string is attached to the ends of the bow and that its curve
is contstant ( always circular - never eliptical.

How would i calculate the radius of the bow?

If i shorten the string by 10cm, how much will the radius change ?  Is the
change linear ? I dont think so.

John

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2002\04\03@073628 by Russell McMahon

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> Now, for the question, one that i could probably solve while i was still
at
> school, but alas, that was many moons ago.
>
> I have a bow ( u know - like in a bow and arrow ).
>
> The lenght of the piece of "wood" and the length of the piece of string is
> known.
> Assume that the string is attached to the ends of the bow and that its
curve
> is contstant ( always circular - never eliptical.
>
> How would i calculate the radius of the bow?
>
> If i shorten the string by 10cm, how much will the radius change ?  Is the
> change linear ? I dont think so.


I suspect you have an infinite number of answers depending on what portion
of a circle you bow makes.

I get the messy answer

   Bow length = 2 x Radius x ASin(String_length/(2 x Radius))

           ASin = inverse Sine (ie angle whose Sine is)

Equally messily this can be put

       String_length/(2 x Radius) = Sin(Bow_length/(2 x Radius))
or    String_length= (2 x Radius) x Sin(Bow_length/(2 x Radius))

In both cases above R is inside and outside a trig function. To solve for it
is messy but it may be a standard identity - anyone?

You could easily solve this for individual cases using a spreadsheet solver
function or similar.

Useless comment:    When the arc of the circle formed by the bow is small
(bow length << radius) you can approximate sin(angle) = angle (in radians)
and you get the result Bow_length = String length (approximately) which is
obvious. This assumption effectively sets R to approximately infinity which
is not much use to you :-)


       Russell McMahon


PS - I'm deeply upset that you don't subscribe to all my uplifting OT posts
:-)

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2002\04\03@153933 by Lee Jones

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> I have a bow ( u know - like in a bow and arrow ).
>
> The length of the piece of "wood" and the length of the piece
> of string is known.  Assume that the string is attached to the
> ends of the bow and that its curve is contstant ( always
> circular - never eliptical.

> How would i calculate the radius of the bow?

You've got a circular arc with a chord.  Length of the circular
arc, i.e. the bow length, we'll call bowlen.  Length of the bow
string we'll call slen (i.e. the chord).  [I'd use strlen as the
variable name, but it has too may C language overtones. :-) ]

The math changes depending if the arc formed by the bow is less
than or greater than half a circle.  I'll do the less than form
which requires:

   bowlen < (slen * pi) / 2

I'm using  r  for the unknown radius...

[use monospaced font for formulas]

Imagine two radius lines from the intersection points of the
bow and bow string to the center of the circle.  These radius
lines and the chord (slen) form an isoceles triangle.  Split
that triangle in half by adding a perpendicular line from the
center of the chord to the center of the circle.  This forms a
right triangle with known side opposite (slen/2) & hypotenus (r).

The angle in the right triangle is

          ( (slen / 2) )
   arcsin ( ---------- )
          (     r      )

Therefore, the angle formed in the isoceles triangle (i.e. two
radius lines capped by the bow) is

              ( (slen / 2) )
   2 * arcsin ( ---------- )
              (     r      )

This angle allows us to know the portion of a full circumference
which the bow represents.

We know that the circumference of the circle is 2 * pi * r, so
the length of the arc is

            (             ( (slen / 2) )  )
            (  2 * arcsin ( ---------- )  )
            (             (     r      )  )
   bowlen = ( --------------------------- ) * 2 * pi * r
            (           2 * pi            )

Since the wood bow length is a given, we can solve for r as it
is the only unknown.

I don't remember all of my trig formulas any more and don't have
a table of them handy, so I won't attempt to reduce the formula.


> If i shorten the string by 10cm, how much will the radius
> change ?  Is the change linear ? I dont think so.

Given the arcsin fuction, I doubt that it's linear.

                                               Lee Jones

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2002\04\04@025843 by Alan B. Pearce

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> If i shorten the string by 10cm, how much will the radius
> change ?  Is the change linear ? I dont think so.

Well lets take some limit values.

If the string length is 0 then the bow forms a circle on its own.

If the string length is half the bow length, then the bow forms a semi
circle.

If the string length = bow length, then the radius is infinite, as the bow
is a straight line.

This seems to be close to an exponential function to me.

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2002\04\04@031734 by Kevin Blain

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I rekon the bow forms a catenary curve, not a circle.


Regards, Kevin

{Quote hidden}

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2002\04\04@043726 by Russell McMahon

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Re 2 people's comments

> I rekon the bow forms a catenary curve, not a circle.

As I understand the problem he defined the box shape to be part of a circle.
I'm not sure that it is a real boz that he is talking about - maybe just an
illustrative name.

> This seems to be close to an exponential function to me.


My formula (possibly wrong) was

   Bow length = 2 x Radius x ArcSin(String_length/(2 x Radius))

Which, unfortunately is somewhat hard to solve for Radius (for me anyway
:-) ).


       RM

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2002\04\04@142345 by Andrew Warren

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Alan B. Pearce <PICLISTspamKILLspammitvma.mit.edu> wrote:

> If the string length is half the bow length, then the bow forms a semi
> circle.

   Only if pi=2.

   -Andy

=== Andrew Warren -- .....aiwKILLspamspam.....cypress.com
=== Principal Design Engineer
=== Cypress Semiconductor Corporation
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=== Opinions expressed above do not
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=== Cypress Semiconductor Corporation

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2002\04\05@023533 by Alan B. Pearce

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>> If the string length is half the bow length, then the bow forms a semi
>> circle.

>    Only if pi=2.

Whoops, but everyone got the point about my taking limiting values :)

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