Turning on snowboard or surfboard

  • Thread starter Thread starter evnmorfun
  • Start date Start date
  • Tags Tags
    Turning
AI Thread Summary
When a rider leans at a 45-degree angle during a turn, the g-force experienced can be calculated using the principles of physics, specifically through the relationship between bank angle, velocity, and turn radius. The effective gravitational acceleration vector must align with the board to maintain balance, and for a 45-degree bank angle, the resultant force felt is approximately 1.4g. The calculation for g-force involves determining the acceleration using the formula a = (velocity)^2 / radius of turn, and then dividing by the gravitational constant. This analysis applies not only to snowboarding but also to surfing and other board sports. Understanding these dynamics is crucial for optimizing performance and safety while riding.
evnmorfun
Messages
2
Reaction score
0
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 independant 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 ?
 
Physics news on Phys.org
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.
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top