Subject:
|
Re: Here's looking at Euclid
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Tue, 1 Aug 2000 18:28:53 GMT
|
Viewed:
|
200 times
|
| |
|
|
In lugnet.off-topic.geek, David Eaton writes:
> In lugnet.off-topic.fun, John Gramley writes:
> > In lugnet.off-topic.fun, Dave Schuler writes:
> > > Since I haven't had geometry in a zillion years, I'm ill-equipped to solve a
> > > little dilemma I've run across, and I thought a few of the more math-minded
> > > among us might be able to help. Here goes:
> > >
> > > Given two points on a circle, can one compute the diameter of the circle • if
> > > the distance between the two points is known? How would one do so?
> > > In the name of mercy, please keep it simple in accordance with my mathless
> > > brain, but I appreciate any help you folks can give!
> >
> > [x-post and f.u.t. to off-topic.geek since this could get complicated]
> >
> > Well, I have a B.A. in math, but I think this should be a simple enough
> > answer: you can't.
> >
> > The main explanation is that it generally takes three distinct points to
> > determine a unique circle. You can get away with two distinct points only if
> > you know something else about them, like that they're endpoints of the
> > diameter.
>
> Hmm... That was my first instinct reaction, however, the thought occurred that
> perhaps what is known about the two points is their distance apart as measured
> along the circumfrence of the circle? (assuming closest distance, but furthest
> would work as well) Hmm... I think in that case, it is possible...
>
> Let's say you've got x1,y1 and x2,y2, and you know the distance between the
> two points along the circumference is L. The smallest circle possible is if
> the center of the circle is at (x1+x2)/2,(y1+y2)/2. Further, we know that both
> points are equidistant from the center of the circle, so the center of the
> circle MUST be on the line:
> (Forgive me if I made an error in algebra)
> Next, construct a function F(x,y) that takes a point (the center of the
> circle) and calculates the partial circumference between the given two points
> (assume closest distance again). I'm bored, but I'm not bored enough to figure
> that one out without either a geometry book or a calc book (Not really sure
> I'd NEED a calc book... maybe to solve it... dunno). Anyway, it's possible to
> find that equasion.
Ok, I was more bored than I thought (assuming degrees, not radians):
F(x,y)=90*pi*(sqrt((x1-x)^2+(y1-y)^2))*sin^-1(((sqrt((x1-x2)^2+(y1-y2)^2))/2)/
(sqrt((x1-x)^2+(y1-y)^2)))
SO, using the formula
L = (all that garbage)
and the equasion I got before:
> y=((x1-x2)/(y1-y2))x+y1+y2+((x1^2+x2^2)/(y1-y2))
you can solve for both x and y using the two equasions, and you'll get x0 and
y0 (the center of the circle). Don't worry, this time I'm REALLY not THAT
bored.
Anyway, then it's just
> 2*sqrt((x1-x0)^2+(y1-y0)^2)
For the diameter.
Anyway, sorry again, but curiosity about whether I could do it or not from
memory was killing me.
DaveE
|
|
Message is in Reply To:
 | | Re: Here's looking at Euclid
|
| (...) Hmm... That was my first instinct reaction, however, the thought occurred that perhaps what is known about the two points is their distance apart as measured along the circumfrence of the circle? (assuming closest distance, but furthest would (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
|
19 Messages in This Thread:
 
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
