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I have a camera moving along a path. The camera position is controlled by an animated Offset value in the Follow Path modifier.

By setting keyframes for the Offset in Graph Editor I can change the camera offset in different sections of the curve, but what I really would like to do is control its speed (the first derivative of the offset). Is it possible to add a speed graph in the Graph Editor?

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  • $\begingroup$ No, Blender doesn't have this feature. $\endgroup$
    – Leander
    Feb 20 at 16:31

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The animation speed is a property of the path, called "Evaluation time". You can find it in the curve's properties. When you animate the path in the constrait menu Blender creates a curve modifier for it in the graph editor. After you delete it you can animate the property as any others, or you can also modify the generator modifier.

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  • $\begingroup$ Thank you. Perhaps I'm missing something, but it seems that "Evaluation time" controls displacement of the object along the path and not the speed. Sorry if I misunderstood your answer. $\endgroup$ Nov 5, 2019 at 3:57
  • $\begingroup$ The object will be located at that percent of the path's lenght if you keep the "Frames" at 100. So if you keyframe the Evaluation time to 0 at the 0th frame and 100 at the 30th frame then the camera will go through the path in 1 second. But if you put 100 at the 300th frame then the trail will last for 10 seconds. So you can adjust the speed with it. $\endgroup$
    – FFeller
    Nov 5, 2019 at 17:03
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    $\begingroup$ Yes, but this is basically what I already can do with the Offset value in the Follow Path modifier. What I'm looking for is a way to set speed explicitly. This is useful when I need to have variable speed along the path. For example, I want 1m/sec between frames A and B. Then 5m/sec between frames B and C. Then 1m/sec again between frames C and D. It's important that the speed is the same between A-B and C-D. Yes, this could be done by trying to adjust the offset slope, but it's inconvenient and imprecise. With an explicit speed graph that would be very straightforward to accomplish. $\endgroup$ Nov 5, 2019 at 21:15

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