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A blaster "bolt" - at least from an E-11 - appears to have a length generally longer than the diameter of a lightsaber blade. If the bolt can be deflected/reflected by a lightsaber, and its length is longer than the diameter of said lightsaber, what would happen to the blaster bolt if the lightsaber were to strike it perpendicularly at the center? Would it split? Be forced into the direction of the swing, like a baseball?

I fear I'm overthinking this, but has that ever happened in any level of canon?

Clarification: So if a were the first point on the bolt which left the blaster, b were a point at roughly 50% of the length, and c were the last point of the blaster bolt to leave the barrel, if a lightsaber were swung and made contact at b, would the entire bolt a-b-c move away together? Considering my understanding is that the blaster projects a gas, I would think only b-c would be deflected while a- would continue on its current path.

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    Given that lightsabers don't pass through other lightsabers, and lightsabers deflect bolts when hitting them, it seems logical that a perpendicular hit would not cut or pass through a bolt. Instead, it would deflect it. – y3sh Dec 10 '15 at 20:25
  • It would split into two like a deadly game of Centipede. – Liesmith Dec 10 '15 at 21:24
  • I don't think bolts are 'solid' when I look at clips (eg this @ 2:40, turn on slow motion) they are reflected like beams and not like a 'stick' would (it would turn around). Although when I look at this starting at 9:00 there are some reflections that could qualify especially at 9:01 there is an interesting flash which did not occur for the previous reflection, but then only one bolt leaves. – bdecaf Dec 10 '15 at 22:09
  • So I guess what we'd need to know first is: whether a blaster bolt is reflected with the bolt a-b-c again, on it's reflected bolt status, is it now a) traveling back towards the sender in a-b-c where the point leading the exit trajectory is again leading the reflection trajectory, or b) does the bolt stay in the same configuration but its direction is reflected, so it is now traveling back towards the sender oriented in the same direction, but having its velocity reversed, returning c-b-a so the first point to return to the sender would be the last point to have left the barrel. – Nate Dec 11 '15 at 13:39

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