In this guide, I’ll be explaining how to create platform jumps that take advantage of the gim’s jumping limit.

Note: this is only for 1x speed and doesn’t calculate coyote jumps

Note #2: this is for approximation only. A gim does not jump exactly 5 tiles high, and neither do they jump exactly 7 tiles wide. The real value has not been calculated yet. This is only for approximating tile placement.

A is the jumping height. B is the horizontal jumping distance. C is the path from the starting point to the end point.

**Update! Thanks to @Blackhole927 , we now have more accurate numbers.**

Max(a) = 4.990959167480469

Max(b) is approx 7 (not accurate)

Cm is the maximum C that a gim can ever have.

Taking note of this, we create an ellipse to show every possible combination of height and distances. (We don’t create a circle because a and b are different in values)

The equation of the ellipse is where b = x and a = y

Now, you just plug in variables into the equation.

Say you want to make a jump that is 5 blocks wide. Set x = 5, and you get 25/49 + (y^2)/25 = 1, which simplifies to y^2 = 25 x 24/49 and then you solve for y, which is 10/7 x sqroot(6), which is approx 3.5. Then you can just create a barrier with a height of 3.5 tiles.

(Note: Above calculation rounds 4.990959167480469 up to 5. for a more precise calculation use the forumla `(b^2)/(4.990959167480469)^2 + (a^2)/(7)^2 = 1`

Thanks for reading!

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