Iwerlipse - Revision history

Omedas: Put section headings back into sentence case

2025-10-17T02:33:12Z

Put section headings back into sentence case

← Older revision		Revision as of 02:33, 17 October 2025
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	In terms of total distance acquired along some direction, forward velocity is thus more effective over long periods of time, where it has time to accelerate to large values, whereas sideways speed is more effective over short periods of time. Optimal straining over many frames then involves straining mostly along the forwards direction to build up a large forward velocity, and then towards the end of the trajectory, transitioning into sideways straining to exploit the large sideways displacement obtainable in a single frame.		In terms of total distance acquired along some direction, forward velocity is thus more effective over long periods of time, where it has time to accelerate to large values, whereas sideways speed is more effective over short periods of time. Optimal straining over many frames then involves straining mostly along the forwards direction to build up a large forward velocity, and then towards the end of the trajectory, transitioning into sideways straining to exploit the large sideways displacement obtainable in a single frame.

	==Update ~~Equations~~==		==Update equations==

	When Mario is in the air and cannot turn, the function <code>update_air_without_turn</code> is responsible for updating Mario's velocity variables from one frame, ''t''-1, to the next, ''t''. Without loss of generality, we can choose coordinates so that Mario's facing angle is zero. In other words, he is facing along the z-axis, while his sideways speed is directed along the x-axis. Mario's forward velocity and position coordinates are then updated via		When Mario is in the air and cannot turn, the function <code>update_air_without_turn</code> is responsible for updating Mario's velocity variables from one frame, ''t''-1, to the next, ''t''. Without loss of generality, we can choose coordinates so that Mario's facing angle is zero. In other words, he is facing along the z-axis, while his sideways speed is directed along the x-axis. Mario's forward velocity and position coordinates are then updated via
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	In deriving the expression for ''z<sub>T</sub>'' we have used that the sum of the first ''T'' integers is equal to the triangle number ''T''(''T''+1)/2, and we have grouped together terms of cos(𝜙<sub>t</sub>). Notice that, since forward straining produces an acceleration that is not reset between frames, earlier frames have a stronger effect on ''z<sub>T</sub>'' than later frames. The effective strength is 1.5(''T''+1-''t''), i.e. 1.5 multiplied by the number of frames remaining (including the current frame), since the forward velocity gained on one frame produces a displacement also on all remaining frames.		In deriving the expression for ''z<sub>T</sub>'' we have used that the sum of the first ''T'' integers is equal to the triangle number ''T''(''T''+1)/2, and we have grouped together terms of cos(𝜙<sub>t</sub>). Notice that, since forward straining produces an acceleration that is not reset between frames, earlier frames have a stronger effect on ''z<sub>T</sub>'' than later frames. The effective strength is 1.5(''T''+1-''t''), i.e. 1.5 multiplied by the number of frames remaining (including the current frame), since the forward velocity gained on one frame produces a displacement also on all remaining frames.

	==Arctan ~~Straining Proof~~==		==Arctan straining proof==

	The displacement vector after ''T'' frames is '''R''' = (''x<sub>T</sub>'' − ''x<sub>0</sub>'') '''x̂''' + (''z<sub>T</sub>'' − ''z<sub>0</sub>'') '''ẑ'''. Taking the dot product of this with the unit vector along the target angle ''ω'' gives the distance moved along this angle		The displacement vector after ''T'' frames is '''R''' = (''x<sub>T</sub>'' − ''x<sub>0</sub>'') '''x̂''' + (''z<sub>T</sub>'' − ''z<sub>0</sub>'') '''ẑ'''. Taking the dot product of this with the unit vector along the target angle ''ω'' gives the distance moved along this angle
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	where we have also rotated the coordinate system back to allow for general facing angles θ. Since tan(''x'') = tan(''x''±180°), this equation has two unique solutions for 𝜙<sub>t</sub> which are given above.		where we have also rotated the coordinate system back to allow for general facing angles θ. Since tan(''x'') = tan(''x''±180°), this equation has two unique solutions for 𝜙<sub>t</sub> which are given above.

	== Iwerlipse ~~Equations~~ ==		== Iwerlipse equations ==

	By writing sin(𝜙<sub>t</sub>) and cos(𝜙<sub>t</sub>) in terms of tan(𝜙<sub>t</sub>), and using the optimal straining relation, we obtain exact expressions for the boundary of the Iwerlipse after ''T'' frames		By writing sin(𝜙<sub>t</sub>) and cos(𝜙<sub>t</sub>) in terms of tan(𝜙<sub>t</sub>), and using the optimal straining relation, we obtain exact expressions for the boundary of the Iwerlipse after ''T'' frames
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	with exponents ''m(T)'' and ''n(T)'' found by minimizing the error. These have been computed up to large ''T'', and can be obtained from a lookup table<ref>[https://drive.google.com/file/d/1xDmgHz878qiAS1ZF1Y5M0M1eiZUdZRSl/view?usp=sharing Showing Iwer's proof and deriving the Iwerlipse expressions.]</ref>. In the limit ''T'' ≫ 1, the exponents converge to ''m'' ≈ 2.520897 and ''n'' ≈ 2.064371, but unlike the integral approximation, the expression never converges to the exact result — there is always some small error.		with exponents ''m(T)'' and ''n(T)'' found by minimizing the error. These have been computed up to large ''T'', and can be obtained from a lookup table<ref>[https://drive.google.com/file/d/1xDmgHz878qiAS1ZF1Y5M0M1eiZUdZRSl/view?usp=sharing Showing Iwer's proof and deriving the Iwerlipse expressions.]</ref>. In the limit ''T'' ≫ 1, the exponents converge to ''m'' ≈ 2.520897 and ''n'' ≈ 2.064371, but unlike the integral approximation, the expression never converges to the exact result — there is always some small error.

	== Quarterstep ~~Penalty~~ ==		== Quarterstep penalty ==

	If a number of quartersteps, ''q'' ∈ {0,1,2,3}, are lost on the final frame, the forward velocity gained on frame ''t'' no longer accumulates forward distance over ''T''+1-''t'' frames, but instead over ''T'' - ''q''/4 + 1 - ''t'' frames. The optimal straining relation then becomes:		If a number of quartersteps, ''q'' ∈ {0,1,2,3}, are lost on the final frame, the forward velocity gained on frame ''t'' no longer accumulates forward distance over ''T''+1-''t'' frames, but instead over ''T'' - ''q''/4 + 1 - ''t'' frames. The optimal straining relation then becomes:
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	sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux].		sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux].

	==See ~~Also~~==		==See also==
	[[Straining]]		[[Straining]]

Omedas: Significantly reduce usage of $tags to reduce chance of "Math extension cannot connect to Restbase" error$

2025-10-17T02:15:16Z

Significantly reduce usage of <math> tags to reduce chance of "Math extension cannot connect to Restbase" error

Show changes

FGSM: change headings to sentence casing

2025-10-16T23:58:57Z

change headings to sentence casing

← Older revision		Revision as of 23:58, 16 October 2025
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	In terms of total distance acquired along some direction, forward velocity is thus more effective over long periods of time, where it has time to accelerate to large values, whereas sideways speed is more effective over short periods of time. Optimal straining over many frames then involves straining mostly along the forwards direction to build up a large forward velocity, and then towards the end of the trajectory, transitioning into sideways straining to exploit the large sideways displacement obtainable in a single frame.		In terms of total distance acquired along some direction, forward velocity is thus more effective over long periods of time, where it has time to accelerate to large values, whereas sideways speed is more effective over short periods of time. Optimal straining over many frames then involves straining mostly along the forwards direction to build up a large forward velocity, and then towards the end of the trajectory, transitioning into sideways straining to exploit the large sideways displacement obtainable in a single frame.

	==Update ~~Equations~~==		==Update equations==

	When Mario is in the air and cannot turn, the function <code>update_air_without_turn</code> is responsible for updating Mario's velocity variables from one frame, <math>t-1</math>, to the next, <math>t</math>. Without loss of generality, we can choose coordinates so that Mario's facing angle is zero. In other words, he is facing along the <math>z</math>-axis, while his sideways speed is directed along the <math>x</math>-axis. Mario's forward velocity, <math>v_t</math>, and position coordinates <math>z_t</math> and <math>x_t</math> are then updated via		When Mario is in the air and cannot turn, the function <code>update_air_without_turn</code> is responsible for updating Mario's velocity variables from one frame, <math>t-1</math>, to the next, <math>t</math>. Without loss of generality, we can choose coordinates so that Mario's facing angle is zero. In other words, he is facing along the <math>z</math>-axis, while his sideways speed is directed along the <math>x</math>-axis. Mario's forward velocity, <math>v_t</math>, and position coordinates <math>z_t</math> and <math>x_t</math> are then updated via
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	Notice that, since forward straining produces an acceleration that is not reset between frames, earlier frames have a stronger effect on the final <math>z</math> position than later frames. The effective strength is <math>1.5(T+1-t)</math>, i.e. 1.5 multiplied by the number of frames remaining (including the current frame), since the forward velocity gained on one frame produces a displacement also on all remaining frames.		Notice that, since forward straining produces an acceleration that is not reset between frames, earlier frames have a stronger effect on the final <math>z</math> position than later frames. The effective strength is <math>1.5(T+1-t)</math>, i.e. 1.5 multiplied by the number of frames remaining (including the current frame), since the forward velocity gained on one frame produces a displacement also on all remaining frames.

	==Arctan ~~Straining Proof~~==		==Arctan straining proof==

	The displacement vector after <math>T</math> frames is		The displacement vector after <math>T</math> frames is
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	Since <math>\tan(x) = \tan(x\pm180^\circ)</math>, this equation has two solutions for <math>\phi_t</math> which are given above.		Since <math>\tan(x) = \tan(x\pm180^\circ)</math>, this equation has two solutions for <math>\phi_t</math> which are given above.

	== Iwerlipse ~~Equations~~ ==		== Iwerlipse equations ==

	By writing <math>\sin(\phi_t)</math> and <math>\cos(\phi_t)</math> in terms of <math>\tan(\phi_t)</math>, and using the optimal straining relation, we obtain exact expressions for the boundary of the Iwerlipse after <math>T</math> frames:		By writing <math>\sin(\phi_t)</math> and <math>\cos(\phi_t)</math> in terms of <math>\tan(\phi_t)</math>, and using the optimal straining relation, we obtain exact expressions for the boundary of the Iwerlipse after <math>T</math> frames:
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	Unlike the integral approximation, this expression never converges to the exact result — there is always some small error.		Unlike the integral approximation, this expression never converges to the exact result — there is always some small error.

	== Quarterstep ~~Penalty~~ ==		== Quarterstep penalty ==

	If <math>q \in {0,1,2,3}</math> quartersteps are lost on the final frame,		If <math>q \in {0,1,2,3}</math> quartersteps are lost on the final frame,
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	sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux].		sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux].

	==See ~~Also~~==		==See also==
	[[Straining]]		[[Straining]]

Aurumaker72: add info about SM64 Lua Redux integration

2025-10-16T17:36:06Z

add info about SM64 Lua Redux integration

← Older revision		Revision as of 17:36, 16 October 2025
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	<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>		<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>
	sm64expert adds arctan straining to MKDasher’s input direction LUA program, making it accessible to everyone for the first time.		sm64expert adds arctan straining to MKDasher’s input direction LUA program, making it accessible to everyone for the first time.

			<div style="font-size:120%; font-weight:bold; margin-top:1em;">16 March 2024</div>
			sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux].

	==See Also==		==See Also==

Aurumaker72: Undo revision 19858 by Aurumaker72 (talk)

2025-10-16T17:30:23Z

Undo revision 19858 by Aurumaker72 (talk)

← Older revision		Revision as of 17:30, 16 October 2025
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	<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>		<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>
	sm64expert adds arctan straining to ~~[https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux]~~, making it accessible to everyone for the first time.		sm64expert adds arctan straining to MKDasher’s input direction LUA program, making it accessible to everyone for the first time.

	==See Also==		==See Also==

Aurumaker72: correct information about the lua script integration

2025-10-16T17:20:43Z

correct information about the lua script integration

← Older revision		Revision as of 17:20, 16 October 2025
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	<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>		<div style="font-size:120%; font-weight:bold; margin-top:1em;">27 November 2021</div>
	sm64expert adds arctan straining to ~~MKDasher’s input direction LUA program~~, making it accessible to everyone for the first time.		sm64expert adds arctan straining to [https://github.com/mupen64/SM64LuaRedux SM64 Lua Redux], making it accessible to everyone for the first time.

	==See Also==		==See Also==

ChouxZi3: pronunciation feels important so I'm moving it to the top

2025-10-16T15:55:48Z

pronunciation feels important so I'm moving it to the top

← Older revision		Revision as of 15:55, 16 October 2025
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	The '''Iwerlipse''' is the range of positions Mario can occupy while he is in the air and cannot turn, assuming that drag is constant. The boundary of the Iwerlipse is reached by '''Arctan Straining''' - the optimal way of air straining that gives the greatest distance along a chosen angle <math>\omega</math> for a given number of frames <math>T</math>. For a single frame <math>T=1</math>, the Iwerlipse is an ellipse with half-width 10 (in the sideways direction) and half-height 1.5 (in the forwards direction). For general <math>T</math>, it resembles an ellipse with half-width <math>10T</math> and half-height <math>3T(T+1)/4</math>, but bulges at the corners, covering more area than an ellipse with the same dimensions. As shown by ''Grassdigger'', the shape is well approximated by a generalised superellipse, and thus ''Pannenkoek2012'' proposed calling this shape the Iwerlipse ~~(pronounced eeverlipse /ˈivɜlɪps/)~~ - a pun on the name of ''Iwer Sonsch'', who first discoverered that Arctan Straining was optimal in April 2018.		The '''Iwerlipse''' (pronounced ee-ver-lipse /ˈivɜlɪps/) is the range of positions Mario can occupy while he is in the air and cannot turn, assuming that drag is constant. The boundary of the Iwerlipse is reached by '''Arctan Straining''' - the optimal way of air straining that gives the greatest distance along a chosen angle <math>\omega</math> for a given number of frames <math>T</math>. For a single frame <math>T=1</math>, the Iwerlipse is an ellipse with half-width 10 (in the sideways direction) and half-height 1.5 (in the forwards direction). For general <math>T</math>, it resembles an ellipse with half-width <math>10T</math> and half-height <math>3T(T+1)/4</math>, but bulges at the corners, covering more area than an ellipse with the same dimensions. As shown by ''Grassdigger'', the shape is well approximated by a generalised superellipse, and thus ''Pannenkoek2012'' proposed calling this shape the Iwerlipse - a pun on the name of ''Iwer Sonsch'', who first discoverered that Arctan Straining was optimal in April 2018.

	==Basics==		==Basics==

Omedas: Explain basic mechanism of Arctan Straining and how it constructs the Iwerlipse. Show Iwer's original proof, and equations for the Iwerlipse. Explain the proper history behind the discoveries.

2025-10-16T15:02:13Z

Explain basic mechanism of Arctan Straining and how it constructs the Iwerlipse. Show Iwer's original proof, and equations for the Iwerlipse. Explain the proper history behind the discoveries.

Show changes

Omedas: Undo revision 19850 by Omedas (talk)

2025-10-13T18:13:40Z

Undo revision 19850 by Omedas (talk)

← Older revision		Revision as of 18:13, 13 October 2025
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			The problem was posed by DeRockProject 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 ~~the TASing/ABC community~~ 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 <math>\text{dyaw}(t,r) = \text{acotan}(0.15r*t) </math>.		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>.

Omedas: Remove explicit mention of DeRockProject, for obvious reasons, and also because many people proposed the question around the same time

2025-10-13T17:55:59Z

Remove explicit mention of DeRockProject, for obvious reasons, and also because many people proposed the question around the same time

← Older revision		Revision as of 17:55, 13 October 2025
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	The problem was posed by ~~DeRockProject~~ 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 the TASing/ABC community 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 <math>\text{dyaw}(t,r) = \text{acotan}(0.15r*t) </math>.		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>.