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How to calculate the point of origin

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Hi, I have a line with a given coordinate (X1,Y1) at one end. If the other end is fixed then I can assume the other end is at (0,0). BUT if the other end is moving or varrying, how can I calculate for the new point of origin, (Xo,Yo)? I would appreciate any reply. Thank you very much.

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Original post by Bong Chua
Hi,

I have a line with a given coordinate (X1,Y1) at one end. If the other end is fixed then I can assume the other end is at (0,0).

BUT if the other end is moving or varrying, how can I calculate for the new point of origin, (Xo,Yo)?

I would appreciate any reply. Thank you very much.
Assuming translation only, the new point of origin is simply the sum of the old point of origin and the current 'other end' point (it's actually the sum of a point and a vector, to be more technically correct).

That seems a little too obvious though, so maybe I'm misunderstanding the question. Perhaps rotation is involved as well? In any case, it will almost certainly be no more complicated than applying a simple affine transform to the original point of origin.

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Dear Jyk,

Thank you for the reply.

I am sorry for my poor explanation. Please consider a standing inclined pencil. At the upper part (eraser), I have (X1,Y1). This point is moving either left, right, down and up, rotation. I dont have problem on this point since I always have a coordinate data.
On the tip of the pencil, I want to find/calculate the coordinate data, (Xn,Yn). This point either moves, forward, backwards, and sideways.

How can I find the coordinates at the tip ?

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Do you mean that you want to simulate the pencil as a rigid body? Like 'how would the tip of the pen move if there was a force F1 on its eraser?'

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Is the body pencil rotating about its own centre, or is just one end of the pencil (X1,Y1) moving around and you want to find out where the other end (X0,Y0) is given that the unknown end is in a fixed relative position to the known end?

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Original post by Enselic
Do you mean that you want to simulate the pencil as a rigid body? Like 'how would the tip of the pen move if there was a force F1 on its eraser?'


- There is no force on the eraser part of the pencil. The movement is like this, for instance. The tip will move first, for example forward, then it stops. Then the eraser part can move either in circular motion, or left or right or up or down. Or,
The tip will move left then it stops. Then the eraser part will start its movement too.
But ofcourse if the tip moves, since the rigid body has a length given, the coordinates at the eraser part changes too.
I forgot to tell that the lenght of the rigid body is also given.

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Original post by Geoff the Medio
Is the body pencil rotating about its own centre, or is just one end of the pencil (X1,Y1) moving around and you want to find out where the other end (X0,Y0) is given that the unknown end is in a fixed relative position to the known end?


-the body of the pencil rotates about its center but it changes its position (Xn,Yn). This is what i want to find out, the coordinates at the tip of the pencil, where i am only given the coordinates at the other end and the length of the pencil.

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Original post by Bob Janova
If the centre doesn't move, then r0 = r1 + 2(rcentre-r1), or (simplified): r0 = 2rcentre-r1

- Kindly bear with me. What is the meaning of r center, r 0, r 1, and r 2?

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Quote:
Original post by Bong Chua
Quote:
Original post by Bob Janova
If the centre doesn't move, then r0 = r1 + 2(rcentre-r1), or (simplified): r0 = 2rcentre-r1

- Kindly bear with me. What is the meaning of r center, r 0, r 1, and r 2?

It's a vector equation (note that there is no r2).

They represent:
rcentre = centre coordinate
r0 = pen tip coordinate
r1 = eraser coordintate

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