Comments on: Spring Physics http://gafferongames.com Glenn Fiedler's Game Development Articles and Tutorials Tue, 31 Jan 2012 03:56:23 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-28006 Glenn Fiedler Fri, 09 Dec 2011 19:05:53 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-28006 Thank you very much! Thank you very much!

]]> By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-28005 Glenn Fiedler Fri, 09 Dec 2011 19:05:38 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-28005 Thanks! Thanks!

]]> By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-28004 Glenn Fiedler Fri, 09 Dec 2011 19:05:23 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-28004 https://ccrma.stanford.edu/CCRMA/Courses/152/vibrating_systems.html http://www.newport.com/Fundamentals-of-Vibration/140234/1033/content.aspx http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx https://ccrma.stanford.edu/CCRMA/Courses/152/vibrating_systems.html
http://www.newport.com/Fundamentals-of-Vibration/140234/1033/content.aspx
http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx

]]> By: Henlus http://gafferongames.com/game-physics/spring-physics/#comment-17978 Henlus Tue, 23 Aug 2011 23:19:18 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-17978 Hi Glenn, I like your article. I'll like to know if you have any idea on how to calculate Mechanical damping of a mass spring damper system? Hi Glenn, I like your article. I’ll like to know if you have any idea on how to calculate Mechanical damping of a mass spring damper system?

]]> By: Peter http://gafferongames.com/game-physics/spring-physics/#comment-474 Peter Sun, 20 Jun 2010 18:29:40 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-474 You might want to explore my App - 4 springs and one mass - www.itunes.com/app/springsnthingslite It uses the Euler-Cromer algorithm which is much better than just Euler because it conserves energy when there is no damping. Cromer tells us to take a half step in approximating the velocity. The full version of Springs'nThings lets you adjust spring damping, pendulum damping, mass, etc. Hope your readers (those with an iPhone/iPad/ITouch) enjoy it. Peter You might want to explore my App – 4 springs and one mass -
http://www.itunes.com/app/springsnthingslite

It uses the Euler-Cromer algorithm which is much better than just Euler because it conserves energy when there is no damping. Cromer tells us to take a half step in approximating the velocity. The full version of Springs’nThings lets you adjust spring damping, pendulum damping, mass, etc.

Hope your readers (those with an iPhone/iPad/ITouch) enjoy it.

Peter

]]> By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-473 Glenn Fiedler Mon, 29 Mar 2010 08:32:22 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-473 That's very interesting david, thanks for the info! That’s very interesting david, thanks for the info!

]]> By: David http://gafferongames.com/game-physics/spring-physics/#comment-472 David Sun, 28 Mar 2010 14:58:08 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-472 I found that advancing all bodies together in each RK4 substep tends to increase the total energy of the system (i.e. explode) more quickly than advancing each body through all RK4 substeps separately. However, advancing each body through all RK4 substeps separately seems to lead to higher systematic errors. For example, the center of mass of a two-mass oscillator tends to float in the direction of the mass which is integrated last. Based on what I found, I personally prefer advancing each body through all substeps separately: it's easier to code and seems to be less prone to explosions in complex systems. I found that advancing all bodies together in each RK4 substep tends to increase the total energy of the system (i.e. explode) more quickly than advancing each body through all RK4 substeps separately. However, advancing each body through all RK4 substeps separately seems to lead to higher systematic errors. For example, the center of mass of a two-mass oscillator tends to float in the direction of the mass which is integrated last.

Based on what I found, I personally prefer advancing each body through all substeps separately: it’s easier to code and seems to be less prone to explosions in complex systems.

]]> By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-471 Glenn Fiedler Wed, 17 Feb 2010 05:45:45 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-471 Fixed now. Fixed now.

]]> By: Glenn Fiedler http://gafferongames.com/game-physics/spring-physics/#comment-470 Glenn Fiedler Tue, 16 Feb 2010 17:17:14 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-470 You are correct, the way I've described it in the article is only correct if the rest length is zero. I'll fix it up. You are correct, the way I’ve described it in the article is only correct if the rest length is zero. I’ll fix it up.

]]> By: Jack http://gafferongames.com/game-physics/spring-physics/#comment-469 Jack Tue, 16 Feb 2010 10:40:06 +0000 http://www.gaffer.org/wordpress/spring-physics/#comment-469 Hi! I want to point out that, actually, your description of equation on springs and dampers is completely wrong. F= -kx .where x is not LENGTH of the string but actually the difference(with sign) between length of relaxed spring and the spring stretched/squeezed(x = deformation). So basically if spring is relaxed the force is F= -k *0 = 0 ; and when lets say its stretched 10 centimetres the F= -k*0,1= -0,1k. so for k =1; normal spring x=0; stretched spring x= 0,1m : F=-0,1 force is negative to the force you use to stretch. squeezed spring x=-0,1m : F=0,1 force is again negative to the force you stretch with, but actually you squeeze, so it want to be relaxed again. BTW. Really great articles! Hi!
I want to point out that, actually, your description of equation on springs and dampers is completely wrong. F= -kx .where x is not LENGTH of the string but actually the difference(with sign) between length of relaxed spring and the spring stretched/squeezed(x = deformation). So basically if spring is relaxed the force is F= -k *0 = 0 ; and when lets say its stretched 10 centimetres the F= -k*0,1= -0,1k.
so for k =1;
normal spring x=0;
stretched spring x= 0,1m : F=-0,1 force is negative to the force you use to stretch.
squeezed spring x=-0,1m : F=0,1 force is again negative to the force you stretch with, but actually you squeeze, so it want to be relaxed again.

BTW. Really great articles!

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