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I have a set of 3D points that I need to "blend" between. Lets say I have points [A,B,C] and I have the amount I want to blend each as [0.5, 0.2, 0.8], how can I blend between these points with the amount I need?

Obviously if I just had 2 points I could use the vector math (A + (B - A) * ratio).

Method 1: I'm currently using a weighted average, which I suppose gives me the weighted center, but I need to try different blending styles to see visually which looks best.

Method 2 What If I did this:

  1. Calculated a difference vector, each 3D point against the default point (I do have a default point)

    A - Default

  2. Then multiplied this difference with the blend amount

    (A - Default) * Amount

  3. And added these up, would that work? What result would I get?

    (A - Default) * A_Amount + (B - Default) * B_Amount + (C - Default) * C_Amount

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    \$\begingroup\$ What do you mean by blending here? Do you want the weighted center of these points? If you just want to blend three points A to three other points B you could use your formula with three different sets of A, B, and ratio. \$\endgroup\$
    – danijar
    Commented Dec 24, 2012 at 9:43
  • \$\begingroup\$ I need to try different types of blending to see visually which works best. I'm currently using a weighted average, which I suppose gives me the weighted center. Are there other types? See edit; \$\endgroup\$
    – Robin Rodricks
    Commented Dec 24, 2012 at 9:47

1 Answer 1

Reset to default
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It is called a barycentre. Here your point is:

P = (A * A_ratio + B * B_ratio + C * C_ratio) / (A_ratio + B_ratio + C_ratio)

Badly, Wikipedia have no dedicaced page for this, so you'll have to understand the explanation of center of mass, which is just a generalization of barycentres applied to physics.

EDIT: your second method is equal to:

P - default

Proof:

((A - def) * A_r + (B - def) * B_r + (C - def) * C_r) / (A_r + B_r + C_r)
= (A * A_r + B * B_r + C * C_r) / (A_r + B_r + C_r) - def * (A_r + B_r + C_r) / (A_r + B_r + C_r)
= P - def
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  • \$\begingroup\$ What would the ratios be if I wanted a perfectly centralized point? 0.5 for all? Or 1 for all? \$\endgroup\$
    – Robin Rodricks
    Commented Dec 24, 2012 at 9:54
  • \$\begingroup\$ It is called an isobarycenter, and is equal to the average of all points \$\endgroup\$
    – Synxis
    Commented Dec 24, 2012 at 9:55
  • \$\begingroup\$ Synxis you are really good in math, but I cannot understand your "P - default" statement or its proof, in simple terms what do you mean? \$\endgroup\$
    – Robin Rodricks
    Commented Dec 24, 2012 at 10:04
  • \$\begingroup\$ P - default: P is the barycenter defined at the top of my answer, and default is the point you've named Default'. In a nushell, your second method is just a 'wrong' version of your first, because the result will be translated by -1 * Default` \$\endgroup\$
    – Synxis
    Commented Dec 24, 2012 at 11:32
  • \$\begingroup\$ No. You cannot set all ratios to 0, because you have to divide by the sum of all ratios, which has to be non-null. \$\endgroup\$
    – Synxis
    Commented Dec 24, 2012 at 13:37

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