Air options
8 posts • Page 1 of 1
Air options
Hi i want to know how air in algodoo works, what is linear,quadratical ...etc. options means.
I was looking for it on forum but i can´t find anything.
Can you help me?
I was looking for it on forum but i can´t find anything.
Can you help me?
- JakubKubo
- Posts: 18
- Joined: Thu Mar 12, 2015 6:27 pm
Re: Air options
AirFriction is just drag forse to polygons. Look at formula in Air Friction's settings
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eanayayo - Posts: 7
- Joined: Fri Sep 11, 2015 7:06 am
Re: Air options
The default equation quoted in game is "-D * (0.01v + 0.001v^2)". D is diameter, v is velocity.
When it says linear, that means the 0.01v term. Because there's only one v, as v goes up by a certain factor, the linear drag will go up by a certain factor. When it says quadratic, that means the 0.001v^2 term. (If you aren't familiar with the notation, v^2 just means v * v.) In a quadratic relationship, if v goes up by a factor, the drag will go up by that factor times itself. For example, the linear drag on a 15 m/s object is half that of a 30 m/s object. But the quadratic drag on a 30 m/s object will be quadruple the drag of that same object moving at 15 m/s.
If I have a sphere with a diameter of 1 moving at 30 m/s, my D is 1, my v is 30. So if I plug that into the equation, I get:
-1 * (0.3 + 0.9) = 1.2 N of drag on that sphere. And indeed, if you create a sphere with diameter (not radius, careful!) of 1, set it moving at 30 m/s, and measure the force, you'll find it gets you 1.2 N of drag.
A simple addition to this is wind speed. In Algodoo, the wind speed is just subtracted from the velocity when drag is calculated. For example, if a ball were moving forward at 10 m/s and the wind was also blowing forwards at 5 m/s, the drag calculations would act as though the ball were moving 5 m/s.
Let me know if any of that was unclear and I'll try and fix it up. I hope this helped some!
When it says linear, that means the 0.01v term. Because there's only one v, as v goes up by a certain factor, the linear drag will go up by a certain factor. When it says quadratic, that means the 0.001v^2 term. (If you aren't familiar with the notation, v^2 just means v * v.) In a quadratic relationship, if v goes up by a factor, the drag will go up by that factor times itself. For example, the linear drag on a 15 m/s object is half that of a 30 m/s object. But the quadratic drag on a 30 m/s object will be quadruple the drag of that same object moving at 15 m/s.
If I have a sphere with a diameter of 1 moving at 30 m/s, my D is 1, my v is 30. So if I plug that into the equation, I get:
-1 * (0.3 + 0.9) = 1.2 N of drag on that sphere. And indeed, if you create a sphere with diameter (not radius, careful!) of 1, set it moving at 30 m/s, and measure the force, you'll find it gets you 1.2 N of drag.
A simple addition to this is wind speed. In Algodoo, the wind speed is just subtracted from the velocity when drag is calculated. For example, if a ball were moving forward at 10 m/s and the wind was also blowing forwards at 5 m/s, the drag calculations would act as though the ball were moving 5 m/s.
Let me know if any of that was unclear and I'll try and fix it up. I hope this helped some!
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ducky21 - Posts: 301
- Joined: Tue Aug 03, 2010 8:58 pm
Re: Air options
ducky21 wrote: D is diameter
D is Density
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eanayayo - Posts: 7
- Joined: Fri Sep 11, 2015 7:06 am
Re: Air options
It specifically says D is the diameter if you double click on air friction. It may be standard to use a big D for density, but I've seen rho more often.
Try to test your claims before you make them - it's the basis of science and a very good habit to get into. Again, if you test it in Algodoo with a ball, you can change the density of the object and the force acting upon it will be the same. The acceleration will change, because acceleration is force/mass, but we only care about the force.
Try to test your claims before you make them - it's the basis of science and a very good habit to get into. Again, if you test it in Algodoo with a ball, you can change the density of the object and the force acting upon it will be the same. The acceleration will change, because acceleration is force/mass, but we only care about the force.
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ducky21 - Posts: 301
- Joined: Tue Aug 03, 2010 8:58 pm
Re: Air options
I only want to find out how to make car what use only air friction
(my car scenes from top view:
http://www.algodoo.com/algobox/details.php?id=114807
or:
http://www.algodoo.com/algobox/details.php?id=110287
)
and how is it with box? if i made two same boxes but one in 0°(slim side) angle and second in 90°(thick side) angle will then "produce" same air friction
Thanks for help, because i don´t understand formula in algodoo (i need to read your answer two times to fully understand )
Sorry for my poor english but i Slovak
(my car scenes from top view:
http://www.algodoo.com/algobox/details.php?id=114807
or:
http://www.algodoo.com/algobox/details.php?id=110287
)
and how is it with box? if i made two same boxes but one in 0°(slim side) angle and second in 90°(thick side) angle will then "produce" same air friction
Thanks for help, because i don´t understand formula in algodoo (i need to read your answer two times to fully understand )
Sorry for my poor english but i Slovak
- JakubKubo
- Posts: 18
- Joined: Thu Mar 12, 2015 6:27 pm
Re: Air options
I'm not sure I entirely understand what you want to do with air friction on your cars, but I made a demonstration on how some of the air friction works. Hopefully it helps, but if not, can you re-explain what you want with the car?
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ducky21 - Posts: 301
- Joined: Tue Aug 03, 2010 8:58 pm
Re: Air options
If box with sides 1x2m is turned at 0° will it have same friction as if turned 90°? because area is two times bigger
- JakubKubo
- Posts: 18
- Joined: Thu Mar 12, 2015 6:27 pm
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