2d geometry problem .... ?

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Re: 2d geometry problem .... ?

Post by laklak » Tue Mar 10, 2020 8:55 pm

Actually, on rereading the OP it's unclear to me if you have the coordinates of P. I read it as you have coordinates for only A, B, and C, but not for arbitrary point P. If you have the four coordinates then you can calculate d directly without reference to angle a. Without coordinates for P you cannot calculate d without additional information.
Yeah well that's just, like, your opinion, man.

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Re: 2d geometry problem .... ?

Post by Brian Peacock » Tue Mar 10, 2020 9:05 pm

JimC wrote:
Tue Mar 10, 2020 8:50 pm
9B has nailed it. You actually don't need the angle you mentioned in your OP, Brian, simply all 4 coordinates, then use 3 different versions of the formula, one for the perpendicular distance to each of the 3 sides. You could, additionally, calculate the angle if you wanted.
The coordinates of A, B, C, and P are known, as is angle a (here, defined from the horizontal ccw). These are the starting conditions, making d a function of a - i.e. when a is different the distance d to the boundary of the triangle is different. See my previous reply to 9B here.
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by NineBerry » Tue Mar 10, 2020 9:12 pm

P and a give you a line. A and B give a line. Calculate X as the point where the two lines intersect. Then just calculate the distance between P and X.

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Re: 2d geometry problem .... ?

Post by Brian Peacock » Tue Mar 10, 2020 9:18 pm

A, B, C, and the point P remain as in Figure B in the OP.

Image
Brian Peacock wrote:
Tue Mar 10, 2020 9:05 pm
... i.e. when a is different the distance d to the boundary of the triangle is different ...
Image
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by NineBerry » Tue Mar 10, 2020 9:18 pm


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Re: 2d geometry problem .... ?

Post by NineBerry » Tue Mar 10, 2020 9:19 pm

Brian Peacock wrote:
Tue Mar 10, 2020 9:18 pm
A, B, C, and the point P remain as in Figure B in the OP.
Brian Peacock wrote:
Tue Mar 10, 2020 9:05 pm
... i.e. when a is different the distance d to the boundary of the triangle is different ...
Image

The selected attachment doesn't exist

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Re: 2d geometry problem .... ?

Post by Brian Peacock » Tue Mar 10, 2020 9:24 pm

Ah. What about now?
Figure_B2.png
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by NineBerry » Tue Mar 10, 2020 9:27 pm

Yeah, see my earlier post

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Re: 2d geometry problem .... ?

Post by Brian Peacock » Tue Mar 10, 2020 9:31 pm

NineBerry wrote:
Tue Mar 10, 2020 9:12 pm
P and a give you a line. A and B give a line. Calculate X as the point where the two lines intersect. Then just calculate the distance between P and X.
Sorry. Not getting it. :dunno:
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by Brian Peacock » Tue Mar 10, 2020 10:13 pm

The distance D between two points in 2d geometry (say, P {x₁, y₁} and X {x₂, y₂} as per 9B's post above) is:

D(P,X) = √((x₂ - x₁)² + (y₂ - y₁)²)

...but I still need to know what d is in my examples in order to find out where the dotted line intersects the triangle's boundary at 9B's X {x₂, y₂}.

What am I missing?

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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by pErvinalia » Wed Mar 11, 2020 12:50 am

Ooh, I love these sorts of problems. I'll have a crack after work today.
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Re: 2d geometry problem .... ?

Post by JimC » Wed Mar 11, 2020 2:00 am

Brian Peacock wrote:
Tue Mar 10, 2020 9:05 pm
JimC wrote:
Tue Mar 10, 2020 8:50 pm
9B has nailed it. You actually don't need the angle you mentioned in your OP, Brian, simply all 4 coordinates, then use 3 different versions of the formula, one for the perpendicular distance to each of the 3 sides. You could, additionally, calculate the angle if you wanted.
The coordinates of A, B, C, and P are known, as is angle a (here, defined from the horizontal ccw). These are the starting conditions, making d a function of a - i.e. when a is different the distance d to the boundary of the triangle is different. See my previous reply to 9B here.
Once you have the coordinates of the 4 points, the issue is fixed, and the angle is not needed as an initial condition; in fact, it can be calculated too, if needed. I'm thinking of making a spreadsheet which, given the 4 initial coordinates, calculates the 3 perpendicular distances, the angles for each, the lengths of the triangle sides, the area of the triangle, and anything else I can think of! :biggrin:
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Re: 2d geometry problem .... ?

Post by Brian Peacock » Wed Mar 11, 2020 2:10 am

You sure know how to have fun! :D

Again, the problem is not finding the shortest (perpendicular) distance from from any point P within the triangle to any of its sides, but calculating the distance from P to the boundary of the triangle given any value for a.

I think the first Figure B was a visually misleading representation as it looked like a perpendicular intersection, so see the revised Figure B here: viewtopic.php?p=1855598#p1855598





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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?

Post by pErvinalia » Wed Mar 11, 2020 5:50 am

Well I've been trying for an hour with triangles and I can't work it out. It requires voodoo maths!
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Re: 2d geometry problem .... ?

Post by pErvinalia » Wed Mar 11, 2020 6:09 am

NineBerry wrote:
Tue Mar 10, 2020 6:51 pm
Connect the dot to all three corners. That creates three triangles. The height of these triangles is the distance of the point to the borders.
The height is the shortest distance. That's not what he is asking.
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