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Iwerlipse: Difference between revisions
Math formatting
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The set of differential equations is: | The set of differential equations is: | ||
dx | <math>\frac{dx}{dt} = v_x\cos[f(t)]</math> | ||
d^2y | <math>\frac{d^2y}{dt^2} = a\sin[f(t)] - D </math> | ||
''f(t)'' is the function for a SM64 player's joystick tilt in terms of time. | |||
''a'' is forward acceleration, and is usually defined in game as 1.5. | |||
''v<sub>x</sub>'' is sideways velocity and is usually defined in game as 10. | |||
''D'' is drag, which usually can be values: +2.35, +0.35, 0, -0.35, -1.35. For TASing purposes, where high speed is always ideal, ''D'' is most commonly set to -1.35, for cases when Mario's forward speed is above 32. | |||
D is drag, which usually can be values: +2.35, +0.35, 0, -0.35, -1.35. For TASing purposes, where high speed is always ideal, D is most commonly set to -1.35, for cases when Mario's forward speed is above 32. | |||
The problem was posed by jongyon7192p as a curiosity, and it became relevant to the ABC challenge a few times before being solved through some other means. | The problem was posed by jongyon7192p as a curiosity, and it became relevant to the ABC challenge a few times before being solved through some other means. | ||
Iwer stated on April 2018 that optimal straining for the Air_no_Turn action is dyaw(t,r) = acotan(0.15r*t). | Iwer stated on April 2018 that optimal straining for the Air_no_Turn action is <math>\text{dyaw}(t,r) = \text{acotan}(0.15r*t) </math>. | ||
On October 2019, trex21415 provided a full analytical proof for Iwer's claim, using a generalized 2nd-order Euler-Lagrange equation. | On October 2019, trex21415 provided a full analytical proof for Iwer's claim, using a generalized 2nd-order Euler-Lagrange equation. | ||
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By plugging Iwer and Trex's equation to solve for x and y, we obtain the parametric equations, the solution to the Iwerlipse: | By plugging Iwer and Trex's equation to solve for x and y, we obtain the parametric equations, the solution to the Iwerlipse: | ||
x = | <math>x = \pm \frac{v_xnt}{\sinh(t)} </math> | ||
y = | <math>y = v_in + \frac{a}{2}n^2\bigg(\coth(t)-\frac{t}{\sinh^2(t)}-\frac{D}{a}\bigg)</math> | ||