Euclidean Distance?

[SteveChandler]

I know that to measure distance between a set of points the equation is SQRT((x2-x1)^2+(y2-y1)^2)

I want to double the length of a line a-b for a new line a-c. How do I solve for c? Is there something in vi.lib?

[SynergyMike]

This stuff was a looonnng time ago for me, but assuming a is common, wouldn't c just be 2b-a (calculated on a per x and y basis), without needing the squares and roots? 

Michael

[SteveChandler]

That might work but I think it falls apart with angled lines.

[altenbach]

Easiset would be to use complex representation for the points:

 

[majoris]

So you have two points. A and B. They make a line segment AB and have length ||AB|| =  sqrt ( (b_x - a_x) ^ 2 + (b_y - a_y ) ^ 2 ).

You want to make a second point C such that ||AC|| = 2 ||AB||.

Then you say some nonesense about angled lines. Solving for C seems pretty easy to me.

Just compute C = 2 * ( B - A ) + A. Where B and A are points (i.e. clusters with an X numeric and Y numeric).

Also, there is a function in IMAQ called IMAQ Point Distances which find N - 1 distances between a list of length N points.

[altenbach]

You did not really specify what datatype "points" are. (in my book I always use complex in 2D :))

The solution by majoris is basically the same as mine, except for the datatype. For the special case of "doubling", we can go even simpler. 😄

 

[Darin.K · Accepted solution]

[SteveChandler]

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@majoris wrote:

So you have two points. A and B. They make a line segment AB and have length ||AB|| =  sqrt ( (b_x - a_x) ^ 2 + (b_y - a_y ) ^ 2 ).

Then you say some nonesense about angled lines. Solving for C seems pretty easy to me.

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Well maybe it is nonsense. Other than that thanks for your answer.

Who to give the accepted solution to? Darins was the smallest and simplest. In fact it is so simple that I regret asking

[SynergyMike]

> That might work but I think it falls apart with angled lines.

Now why would you post something that proves my point?  🙂  (Sqrt of 200 is 14.142...)