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I'm trying to make a character in 2.5 platformer (in UDK) to "climb" a giant tree trunk by walking on a spiral staircase enveloped around the tree. When character goes right the tree rotates thru matinee sequence so it seems that the character is moving while in reality it is the tree that moves.

I connected the matinee sequence playrate to the velocity of the character and its all good as long the character just moves left or right. When it jumps though, the velocity still affects the playrate - it should not as character moves up/down, not right/left.

How do I set it up in Kismet so I get a float variable with velocity only in the X plane (horizontal)?

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    \$\begingroup\$ I'm not familiar with Kismet, but perhaps there's a way to read only the X coordinate of the velocity vector. Because that's exactly what you're looking for. \$\endgroup\$ – Marton Oct 4 '12 at 13:16
  • \$\begingroup\$ Getting the velocity vector gives me nothing as it doesn't tell what is the velocity at the X plane. \$\endgroup\$ – anna1987 Oct 4 '12 at 15:20
  • \$\begingroup\$ @anna1987: Can't you just use the X component of the velocity vector? \$\endgroup\$ – Nic Foster Oct 4 '12 at 18:49
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    \$\begingroup\$ @anna1987 A 2D or 3D velocity vector's X component is exactly the velocity in the X plane. If you can't access the Vector.X property's value (again, I'm not sure how Kismet works), then just multiply the original velocity vector with a new Vector(1,0), where 1 is the X component and Y is 0. This is gonna give you the velocity in the X plane as a vector. \$\endgroup\$ – Marton Oct 4 '12 at 20:16
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Option 1. A 2D or 3D velocity vector's X component is exactly the velocity in the X plane.

Option 2. If you can't access the Vector.X property's value (again, I'm not sure how Kismet works), then just multiply the original velocity vector with a new Vector(1,0), where 1 is the X component and Y is 0. This is gonna give you the velocity in the X plane as a vector.

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"@anna1987 A 2D or 3D velocity vector's X component is exactly the velocity in the X plane. If you can't access the Vector.X property's value (again, I'm not sure how Kismet works), then just multiply the original velocity vector with a new Vector(1,0), where 1 is the X component and Y is 0. This is gonna give you the velocity in the X plane as a vector. – Marton"

Of course it is! I'm such a retard stupid blonde! Thanks guys! xoxo

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  • \$\begingroup\$ Haha okay, I added my comment as an answer, so you can accept it and close the question. \$\endgroup\$ – Marton Oct 5 '12 at 9:26

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