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    Anchor point

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    • iTap DevelopmentI
      iTap Development @Aidan_Fire
      last edited by iTap Development

      @Aidan-Oxley hey, so I’m just coming back to this(I got a 3d printer so I’ve been busy with that!);
      What do you need the “initial angle” for?
      And I’m not sure as to the purpose of the behaviors in the screen shot.0_1509768978624_1A339C64-88B6-41E1-8C8C-A6109AC1CEDC.png

      Jack8680J Aidan_FireA D 3 Replies Last reply Reply Quote 0
      • Jack8680J
        Jack8680 @iTap Development
        last edited by Jack8680

        @iTap-Development those behaviours are Pythagoras' theorem. The result from the square root is the distance from the centre to the anchor point.

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        • Aidan_FireA
          Aidan_Fire @iTap Development
          last edited by

          @iTap-Development Jack is right. This is how you work out the diagonal distance between two points if you know their coordinates.

          iTap DevelopmentI 1 Reply Last reply Reply Quote 0
          • D
            Deeeds @iTap Development
            last edited by

            @iTap-Development A hacky, tacky way to achieve the goal:

            Create a dummy object that you'll use as a virtual anchor point. Use the physics weld joint to bind your rotating object to the dummy object at the position you want relative to the virtual anchor point. Then rotate the dummy object. This should give you the result you want, but will ruin anything else regarding physics.

            But I agree. The anchor point should be respected by the rotate behaviour.

            iTap DevelopmentI 1 Reply Last reply Reply Quote 0
            • iTap DevelopmentI
              iTap Development @Deeeds
              last edited by

              @Deeeds that would work kinda, but yeah it wouldn’t work well in some ways.

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              • iTap DevelopmentI
                iTap Development @Aidan_Fire
                last edited by

                @Aidan-Oxley @Jack8680 oh duh! So you are just multiplying the x and y by them selves for the x and y distance to the other point?
                I think I always just added the actual x and y distances.

                Aidan_FireA 1 Reply Last reply Reply Quote 0
                • Aidan_FireA
                  Aidan_Fire @iTap Development
                  last edited by Aidan_Fire

                  @iTap-Development You know Pythagora’s theorem? If you have a right angle triangle, a is the length of one side, b is the length of the other and c is the length of the hypotenuse (the longest one, diagonal if a is x and b is y), then c² = a² + b². What I was trying to work out is the diagonal distance between two points. I can easily get the x and y distance between them, but I wanted diagonal.

                  iTap DevelopmentI 2 Replies Last reply Reply Quote 0
                  • iTap DevelopmentI
                    iTap Development @Aidan_Fire
                    last edited by

                    @Aidan-Oxley sorry! I was confusing with something else in my project 🤦‍♂️ I’ve only got a few hours of sleep every night for a week due to my 3d printer addiction! so my brain doesn’t work....

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                    • iTap DevelopmentI
                      iTap Development @Aidan_Fire
                      last edited by

                      @Aidan-Oxley so why are you multiplying the sine and cosine angles by the diagonal distance?

                      Aidan_FireA 1 Reply Last reply Reply Quote 0
                      • Aidan_FireA
                        Aidan_Fire @iTap Development
                        last edited by

                        @iTap-Development I don’t remember lol. There is a reason though. It’s probably one of those formulas like hypotenuse = sin(θ) × opposite.

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