Turning on snowboard or surfboard

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SUMMARY

The discussion centers on calculating g-forces experienced by riders on snowboards and surfboards during turns. It establishes that the g-force felt by a rider is influenced by the bank angle, with a 45-degree angle resulting in a resultant force of 1.4g. The conversation emphasizes that the effective gravitational acceleration vector must pass through the board to maintain balance, and the relationship between velocity, radius of curvature, and g-forces is defined using the formula a=(velocity)^2/(radius of your turn).

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evnmorfun
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In a turn, a rider might be leaning over at around 45 degrees at times. What is the g-force at that point going down into his feet ? I'm thinking that it is essentially independent of the speed and turn radius. Also, I'm thinking the rider is close enough to standing straight - no bent knees or hips, arms near to by his or her sides. What about getting really low - say 30 degrees ? Is it a simple sin cos or tan calculation ?
 
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the force due to gravity is his mass times the local gravitational field strength- neither depend on any angles. In snowboarding the rider himself doesn't lean, you still keep your weight more or less over the board (maybe a little in a fast tight turn), mostly you angle the board so that the "bloated" ends bend upwards slightly, and then you trace a curved path.
 
That's not correct. You can calculate the g-force (acceleration) required for a balanced turn for a snow board, surf board - even an airplane, knowing only the bank angle.

For a snowboard and surfboard, in order to not fall over, the resultant effective gravitational acceleration vector of your center of gravity must pass through your board. Ie, if you lean to far to the right, you fall over to the right. So to balance the forces, you draw a diagram showing the resultant vector pointing from your CoG to the board, the gravitational force vector straight down, and a third vector horizontal into the turn. You have all three angles and the magnitude of one side, so you can solve for the magnitude of the other sides.

For a 45 degree bank angle, you have a 45/45/90 right triangle, so the horizontal and vertical components are both 1g and the resultant force (the one you "feel") is 1.4g.
 
I think about this when I'm riding my boards all the time!

How many g's you are pulling is dependent on your velocity and radius of curvature.

Take the constant velocity circular motion equation for normal accelleration:

a=(velocity)^2/(radius of your turn)

to find how many g's you pull divide 'a' by 'g', gravity constant.
 
Thanks , Russ , I definitely like yours the most.
 

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