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hyperPad

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  3. Feature Request: Direction Wheel for Apply Force

Feature Request: Direction Wheel for Apply Force

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  • Aidan_FireA Aidan_Fire

    @Deeeds You could try use propel object if you want an object to have a force applied in the direction its facing. You just have the object propel itself. Say, to propel upwards (when the object is rotated at zero), target A is the object with the y anchor changed to something like 45%, and target B is the same object with the anchor left default.

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    Deeeds
    wrote on last edited by
    #3

    @Aidan-Oxley Not nearly the same.

    Not what I'm asking.

    Not the desired result.

    How does one calculate the force to be applied in any given direction?

    eg: direction = 243 degrees

    If zero degrees is up and north (I'm assuming that's how hyperPad does things... could be wrong, it might be up)

    And the force for 0 degrees is x= 0 and y= 100

    How does one rotate that force to 243 degrees?

    I do NOT know maths.

    Aidan_FireA 1 Reply Last reply
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    • D Deeeds

      @Aidan-Oxley Not nearly the same.

      Not what I'm asking.

      Not the desired result.

      How does one calculate the force to be applied in any given direction?

      eg: direction = 243 degrees

      If zero degrees is up and north (I'm assuming that's how hyperPad does things... could be wrong, it might be up)

      And the force for 0 degrees is x= 0 and y= 100

      How does one rotate that force to 243 degrees?

      I do NOT know maths.

      Aidan_FireA Offline
      Aidan_FireA Offline
      Aidan_Fire
      wrote on last edited by
      #4

      @Deeeds X force = diagonal force Γ— cos(angle)
      Y force = diagonal force Γ— sin(angle)
      This is assuming that 0Β° is to the right, 90Β° is up, 180Β° is left, 270Β° is down (conventional angles). But hyperPad angles don’t work like that. @Hamed do you think that the angle system could be changed a little to at least make anticlockwise positive? Just to make working with trig formulas easier?

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      • HamedH Offline
        HamedH Offline
        Hamed
        Admin
        wrote on last edited by
        #5

        @Aidan-Oxley unfortunately this was a design choice early on. Does adding 90 degrees then multiplying by one not work?

        Aidan_FireA 1 Reply Last reply
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        • Aidan_FireA Aidan_Fire

          @Deeeds X force = diagonal force Γ— cos(angle)
          Y force = diagonal force Γ— sin(angle)
          This is assuming that 0Β° is to the right, 90Β° is up, 180Β° is left, 270Β° is down (conventional angles). But hyperPad angles don’t work like that. @Hamed do you think that the angle system could be changed a little to at least make anticlockwise positive? Just to make working with trig formulas easier?

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          Deeeds
          wrote on last edited by
          #6

          @Aidan-Oxley Yes! I was already having trouble before realising up was 0. I'm also used to 0 being right. Most of the time. I think. Or something like that.

          So how do you calculate in this space?

          Aidan_FireA 1 Reply Last reply
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          • HamedH Hamed

            @Aidan-Oxley unfortunately this was a design choice early on. Does adding 90 degrees then multiplying by one not work?

            Aidan_FireA Offline
            Aidan_FireA Offline
            Aidan_Fire
            wrote on last edited by
            #7

            @Hamed It’s fine, it’s still possible to do everything, just needs some more maths. How come you chose clockwise as positive though? To work completely with physics formulas, I have to reverse the direction and then add or subtract by 90Β° (I keep forgetting exactly how I do this, but I have done it).

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            • D Deeeds

              @Aidan-Oxley Yes! I was already having trouble before realising up was 0. I'm also used to 0 being right. Most of the time. I think. Or something like that.

              So how do you calculate in this space?

              Aidan_FireA Offline
              Aidan_FireA Offline
              Aidan_Fire
              wrote on last edited by
              #8

              @Deeeds You can get it to work by having 90 - object angle. Everything will work fine even though the numbers you get might look completely wrong. Getting the angle values to actually look right as text isn’t that easy.

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              • Aidan_FireA Aidan_Fire

                @Deeeds You can get it to work by having 90 - object angle. Everything will work fine even though the numbers you get might look completely wrong. Getting the angle values to actually look right as text isn’t that easy.

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                Deeeds
                wrote on last edited by
                #9

                @Aidan-Oxley Sorry, talk to me like I don't know what you're talking about.

                Step through the process, please. I really don't know what you mean.

                Where do I do the 90 reduction?

                Aidan_FireA 1 Reply Last reply
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                • D Deeeds

                  @Aidan-Oxley Sorry, talk to me like I don't know what you're talking about.

                  Step through the process, please. I really don't know what you mean.

                  Where do I do the 90 reduction?

                  Aidan_FireA Offline
                  Aidan_FireA Offline
                  Aidan_Fire
                  wrote on last edited by Aidan_Fire
                  #10

                  @Deeeds Add a Get Rotation behaviour and select your object. Add a Subtract Values behaviour, in the top part put in 90 and in the bottom part put in the output of Get Rotation. Now you can use this result instead of object rotation with the other formulas I said earlier to get an angled force πŸ™‚

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                  • Aidan_FireA Aidan_Fire

                    @Deeeds Add a Get Rotation behaviour and select your object. Add a Subtract Values behaviour, in the top part put in 90 and in the bottom part put in the output of Get Rotation. Now you can use this result instead of object rotation with the other formulas I said earlier to get an angled force πŸ™‚

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                    Deeeds
                    wrote on last edited by
                    #11

                    @Aidan-Oxley Cheers. I'll try. Math. My eyes glaze over and my hand reaches for a shot of whiskey every.single.time

                    Aidan_FireA 1 Reply Last reply
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                    • D Deeeds

                      @Aidan-Oxley Cheers. I'll try. Math. My eyes glaze over and my hand reaches for a shot of whiskey every.single.time

                      Aidan_FireA Offline
                      Aidan_FireA Offline
                      Aidan_Fire
                      wrote on last edited by Aidan_Fire
                      #12

                      @Deeeds MAAAAATHS!!! 😍😍😍😍 πŸ€ͺ

                      iTap DevelopmentI D 3 Replies Last reply
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                      • Aidan_FireA Aidan_Fire

                        @Deeeds MAAAAATHS!!! 😍😍😍😍 πŸ€ͺ

                        iTap DevelopmentI Offline
                        iTap DevelopmentI Offline
                        iTap Development
                        wrote on last edited by
                        #13

                        @Aidan-Oxley πŸ‘πŸ€£πŸ€£

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                        • Aidan_FireA Aidan_Fire

                          @Deeeds MAAAAATHS!!! 😍😍😍😍 πŸ€ͺ

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                          Deeeds
                          wrote on last edited by
                          #14

                          @Aidan-Oxley What is diagonal force?

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                          • Aidan_FireA Aidan_Fire

                            @Deeeds MAAAAATHS!!! 😍😍😍😍 πŸ€ͺ

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                            Deeeds
                            wrote on last edited by
                            #15

                            @Aidan-Oxley I put in what I'd call the "desired parallel force" since it's the amount along any one axis that I like... and that worked. Which is 10, in my case.

                            Does this mean that if you want something to move at 45 degrees at the same rate you'd like at one of the parallels, that you do this:

                            X travel desired force: x=10, y=0
                            Y travel desired force: x=0, y=10

                            And therefore diagonal is x=10, y=10 ?

                            I'd have thought it would be:

                            x=5, y=5

                            And that this would be called the diagonal force of x, and the diagonal force of y...

                            Oh wait. I think I figured it out.

                            Diagonal force is the combined force of X and Y to get the desired result???

                            Right?

                            I have no idea.

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                            • D Deeeds

                              @Aidan-Oxley I put in what I'd call the "desired parallel force" since it's the amount along any one axis that I like... and that worked. Which is 10, in my case.

                              Does this mean that if you want something to move at 45 degrees at the same rate you'd like at one of the parallels, that you do this:

                              X travel desired force: x=10, y=0
                              Y travel desired force: x=0, y=10

                              And therefore diagonal is x=10, y=10 ?

                              I'd have thought it would be:

                              x=5, y=5

                              And that this would be called the diagonal force of x, and the diagonal force of y...

                              Oh wait. I think I figured it out.

                              Diagonal force is the combined force of X and Y to get the desired result???

                              Right?

                              I have no idea.

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                              D Offline
                              Deeeds
                              wrote on last edited by
                              #16

                              @Aidan-Oxley @Hamed

                              THANK YOU BOTH!!

                              This is now working. I have objects I can rotate and they apply force correctly.

                              Feature request still stands! It would be awesome if this guff could be abstracted away to where it belongs, in the dustbin with the casio calculator watches of the 1980's.

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                              With your input, this post could be even better πŸ’—

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