As a continuation of my last post, I thought I'd do a few calculations to figure out just how fast a snowboarder should be able to go.
Obviously, the steeper the slope the higher speeds that are possible. Because my last post was motivated by the discussion of avalanches, I decided to do the calculations using a slope of 38 degrees. That's the incline that Wikipedia says is the most likely to have avalanches induced by human activity (snowboarding, skiing, hiking, etc.)
When you bomb a hill on a snowboard (that's what my little brother calls it when you head full tilt down a slope), you have basically three forces to deal with.
- Gravity pulling you down. Of course, the ground below you gets in the way, preventing you from falling straight to the center of the earth, but some of the force of gravity ends up getting directed along the face of the hill, pulling you along the surface and down to the lifts so you can start back up again.
- Air Drag pushing back against you. The thing about drag is it increases much more as you speed up than you might expect. If you double your speed, the drag quadruples.
- Friction. Unlike drag, friction stays relatively constant regardless of your speed down the hill. Although, this isn't exactly true, it's a good approximation. All kinds of things affect the amount of friction you encounter on snow, from the temperature of the snow to it's texture. It's likely that sliding friction changes at least a little as you speed up, but the change is probably pretty small, at least on the firmly packed snow where speed records are set.
Gravity pulls you down the hill, while drag and friction are directed back up the hill. As you build up speed down a slope, the forces of gravity and friction stay constant (provided the slope doesn't change). The drag, on the other hand increases until the forces pulling you down and the forces pushing back up balance out. At that point, you've reached terminal velocity for the hill.
For those of you who like equations, terminal velocity occurs when
Force of gravity = Drag Force + Frictional Force
Skip to the bottom of the post to see it in even more mathy terms.
To cut to the chase, if you assume a 70 kilogram (154 Lbs) snowboarder standing straight up on their way down the slope (giving them a frontal area I estimate to be about 0.5 square meters), with a drag coefficient of about 1 (which Wikipedia tells me is typical of skiers, and is probably about right for snowboarders), a friction coefficient of 0.04 (typical for waxed surfaces on packed snow), and a 38 degree slope, you get a top speed of about . . . 80 mph (129 kph).
Go into a racing tuck, put on a racing helmet and some slick racing clothes, and you can pretty easily push that up over 125 mph (200 kph), which is coincidentally just about the top speed record on a snowboard.
In fact, the fastest speed possible on a 38 degree slope is probably much higher. I made some pretty conservative guesses in doing these calculations, and didn't consider aerodynamics much at all. In fact, the skiing speed record of about 156 mph (252.4 kph), which was set under similar conditions that I used in my calculations, shows that there's lots of room to improve snowboard speed records.
So why haven't snowboard speed records equaled skiing speeds already? That's a subject for another post. I'll tell you what I think about it some other day.
The mathy bits.
Here's the equation I started with to calculate a snowboarder's terminal velocity
m*g*sin(38 degrees)= u*m*g*cos(38 degrees) + (p*A*Cd*v^2)/2
where
'*' is the multiplication symbol and '/' is the division symbol
'm' is the mass of the snowboarder
'g' is the acceleration due to gravity
'p' is the density of air
'Cd' is the drag coefficient
'u' is the friction coefficient
'A' is the frontal area of the snowboarder
And 'v^2' means velocity squared, or v*v
You can go through and rearrange this equation with a little algebra to solve for the velocity.
Sunday, October 7, 2007
Speed Snowboarding - 125 mph and Beyond
Posted by Buzz Skyline at 9:15 AM