2d geometry problem .... ?
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Re: 2d geometry problem .... ?
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 .... ?
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.JimC wrote: ↑Tue Mar 10, 2020 8:50 pm9B 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.
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
Re: 2d geometry problem .... ?
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 .... ?
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 ...
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
Re: 2d geometry problem .... ?
Brian Peacock wrote: ↑Tue Mar 10, 2020 9:18 pmA, 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 ...
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Re: 2d geometry problem .... ?
Ah. What about now?
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
Re: 2d geometry problem .... ?
Yeah, see my earlier post
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Re: 2d geometry problem .... ?
Sorry. Not getting it.
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?
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?
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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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?
Ooh, I love these sorts of problems. I'll have a crack after work today.
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Re: 2d geometry problem .... ?
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!Brian Peacock wrote: ↑Tue Mar 10, 2020 9:05 pmThe 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.JimC wrote: ↑Tue Mar 10, 2020 8:50 pm9B 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.
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Re: 2d geometry problem .... ?
You sure know how to have fun!
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
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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"It isn't necessary to imagine the world ending in fire or ice.
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: 2d geometry problem .... ?
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 .... ?
The height is the shortest distance. That's not what he is asking.
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