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My problem is:

I want two Locators to simulate a vector in the following sense:

The first Locator is the base and the second the tip of the vector.

When I move the tip, the base does not move (hence the vector changes).

When I move the base, the vector is unchanged, and therefore the tip (and the Locator) moves.

How can I achieve that?

I have tried storing the previous value of the base and then test if the current value is different. And if so updated the position of the tip. I can however not make that work, when using 'Module'. I suspect that there is a more elegant solution.

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This you look at the other related answers? Such as this:… – C. E. Aug 13 '13 at 10:36
Also related:… – Michael E2 Aug 13 '13 at 15:26
up vote 5 down vote accepted

This seems to be a duplicate but I can't find it :). Meanwhile, you can use second argument of Dynamic.

x = {0, 0}; y = {1, 1}; w = y - x;

                 Locator@Dynamic[x, (x = #; y = x + w;) &],
                 Locator@Dynamic[y, (y = #; w = y - x;) &],
                 Dynamic@Arrow[{x, y}]
                , PlotRange -> 2]

In case of multiple vectors one may want to save space and make code more transparent so we can use extended version of Dynamic second argument to achieve this:

                 Locator@Dynamic[x, {(w = y - x;) &, (x = #; y = x + w) &, None}],
                 Dynamic@Arrow[{x, y}]
               , PlotRange -> 2]

Now we are working only with base, moreover w can be scoped to particular Locator.

There is huge advantage of the second method, well, not exactly the method but the usage of f_start and f_end. You can calculate w once, not all the time you are dragging the Locator.

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+1 I like this answer a lot and a learned something from it! btw. while I researched it I noticed that the exact number of times w is calculated in the second version is twice every time you drag that locator. (Method; first n=0; and then inside the anonymous function Print[n++]). – C. E. Aug 13 '13 at 13:41
@Anon Indeed, I saw this too with Print. Do you know why it is so? – Kuba Aug 13 '13 at 13:45
No! I became very curious but I suppose there is no good answer. It's not what is says in the documentation so I guess that's just "how it is". – C. E. Aug 13 '13 at 13:47
p1save = {0, 0}; p2save = {1, 1};
        If[p1 != p1save, p2 = p2 + p1 - p1save; p1save = p1];
        Graphics[Arrow[{p1, p2}], PlotRange -> {{-5, 5}, {-5, 5}}, AspectRatio -> 1],
 {{p1, p1save}, Locator}, {{p2, p2save}, Locator}

or a little bit more sophisticated, preventing the arrow to overflow the graphics window:

p1save = {0, 0}; p2save = {1, 1};  min = -5; max = 5;
 If[Or @@ ((min > # || # > max) & /@ Flatten[{p1, p2 + p1 - p1save}]),
                                               p1 = p1save; p2 = p2save];
 If[p1 != p1save, p2 = p2 + p1 - p1save; p1save = p1; p2save = p2];
 Graphics[Arrow[{p1, p2}], PlotRange -> {{min, max}, {min, max}},  AspectRatio -> 1],
 {{p1, p1save}, Locator}, {{p2, p2save}, Locator}
share|improve this answer
Looking good. But somehow in the second example if I move the base of the arrow, the head of the arrow moves at constant speed. – Jacob Akkerboom Aug 15 '13 at 21:22

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