MHB Finding Horizontal/Vertical Distance Between Two Points in Scratch 3.0

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To accurately position the bullet in a Scratch 3.0 game, the shooter sprite's barrel must be aligned with the bullet's trajectory based on the sprite's rotation. The X and Y distances from the sprite's center to the barrel's end vary with the angle of rotation, complicating the calculation. When the sprite faces 90º, the distances are 105 pixels horizontally and 45 pixels vertically. To find the distances for other angles, using trigonometric functions or Pythagorean theorem is suggested, as it forms a right triangle. This approach will ensure the bullet appears to exit the gun correctly regardless of the sprite's orientation.
Kizza23
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I am making a Scratch 3.0 game. The shooter sprite is holding a gun slightly off-centre (see images), and I need the bullet to go to the end of the barrel of the gun before traveling forward (as so it would appear the bullet it leaving the gun). The issue is, to do this, I need to find the X and Y distance of the end of the barrel from the centre of the shooter sprite (Scratch doesn't let you use diagonal distances 😡). The X and Y distances change as the sprite rotates.

I know if the shooter sprite is facing 90º (right angle), the X distance to the gun is 105 pixels, and the Y distance is 45 pixels.

The computer always knows what direction the shooter sprite is facing.

What about the distances for all the other angles the shooter sprite is facing? Do I need a special formula?

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Looks like a right triangle is formed in each case you've shown ... couldn't the distance from your set origin to the muzzle (the hypotenuse) be found using Pythagoras?
 

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