Hi, I am not going to dust off my pure maths books after 35 yrs of being out
of school. Anway I dont have them !
 I am sure there is a maths fix to this question.

Q1) If there are eight circles each 25mm diameter,  in circular orbit about
a centre point, (i.e they all are the same distance out from the centre
point) with shortest distance between any two adjacent circles of 9.74mm,
what is the distance from centre point to the centre of one of those circles
?

I am faced with trial and error on a vector prog to solve this otherwise :-(

Many thanks.
Steve

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.

In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
 "Steve"  wrote:

> .... 
> Q1) If there are eight circles each 25mm diameter,  in circular orbit about
> a centre point, (i.e they all are the same distance out from the centre
> point) with shortest distance between any two adjacent circles of 9.74mm,
> what is the distance from centre point to the centre of one of those circles
> ? ....


      This seems to mean that the distance between the centres of two 
adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?

      If so, then consider the triangle whose three vertices are those 
two centres and the main centre of the diagram.  One of its sides has 
length 34.74 mm, the other two sides have the unknown length you want, 
and the angle between these is 45 degrees.  Applying the cosine law to 
that triangle and doing a bit of algebra gives the length you want as

(34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)

            Ken Pledger.

Many thanks, that indeed is so and understood by you.
Now why didnt I think of it that way !
In fact another way I now see is its the hypo of a triangle 22.5deg at
centre of orbit and half of 34.74mm t'other end.
Thanks
Steve

"Ken Pledger"  wrote in message
news:ken.pledger-4334AC.09392731052007@bats.mcs.vuw.ac.nz...
> In article <uSk7i.28232$Ro3.27449@text.news.blueyonder.co.uk>,
>  "Steve"  wrote:
>
> > ....
> > Q1) If there are eight circles each 25mm diameter,  in circular orbit
about
> > a centre point, (i.e they all are the same distance out from the centre
> > point) with shortest distance between any two adjacent circles of
9.74mm,
> > what is the distance from centre point to the centre of one of those
circles
> > ? ....
>
>
>       This seems to mean that the distance between the centres of two
> adjacent circles is  25 + 9.74 = 34.74 mm.  Is that right?
>
>       If so, then consider the triangle whose three vertices are those
> two centres and the main centre of the diagram.  One of its sides has
> length 34.74 mm, the other two sides have the unknown length you want,
> and the angle between these is 45 degrees.  Applying the cosine law to
> that triangle and doing a bit of algebra gives the length you want as
>
> (34.74)/sqrt(2 - sqrt(2))   which is about 45.39 mm   (E. & O. E.)
>
>             Ken Pledger.