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Question by Reject76 · May 19, 2019 at 05:45 PM · scripting beginnerscriptingbasicsforcedirectional

How to add force in Position X direct of travel

Hello

I know I can add this line to add force in the direction of travel to an object with a rigidbody:

 rigidbody.AddForce(rigidbody.velocity.normalized * Time.deltaTime * forceAmount);

However, I need help in finding a way to add force for only X or -X when the object is travelling in either X direction. I.e. if the object is travelling along -X, force should be applied to -X and vice versa for X.

Does anyone know how to manage that? Many thanks!

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Answer by tormentoarmagedoom · May 20, 2019 at 09:40 AM

Hello.

I'm not sure what are you asking, because its like you know the answer but dont see it...

if want add force in X axis, just:

 rb.Addforce(new Vector3(1,0,0);

Now, you can multiply/divide as you want, but this force will have only X component...

Whats the question ?

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Answer by Reject76 · May 21, 2019 at 05:00 PM

Yes, close, but this is what I'm after:


If the object is travelling in -X direction I want to Addforce -5 on the X axis. If the object is travelling in X direction, I want to Addforce 5 on the X axis.

I'm building a pong-style game and the main problem I have is with the ball getting stuck endlessly bouncing back and forth on the Y-axix betweet my upper and lower border walls (player paddles are left and right).

Hence I need something that shoves the ball either left or right (i.e. X or -X) when the ball hits something on the playfield (wall, paddle or other object).

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