Minimum Slope Required or How Steep is Unsteep

ski_it

Zamboni tailings got me to wondering. I'm sure this has come up before and is well known in the industry, but what is considered the minimum slope needed to yield effortless motion down a snow covered pitch?

lotsoskiing

Meadowlark at Smuggs....

NewEnglandSkier13

It depends on the snow conditions.

rickbolger

probably also depends on the coefficient of friction for any given snow sliding device, which would be affected by surface area, surface roughness, lubricity, and probably some other stuff

ciscokid

Wow that’s deep

rickbolger

Deep -- didn't think of that! Snow depth could affect friction. I suppose that falls under "conditions" as Sam mentioned above

riverc0il

NewEnglandSkier13;c-35151 wroteIt depends on the snow conditions.

Plus ski bases. Mine have a lot of gouges and are never waxed. I'm always shocked how easily skis slide underfoot on flat surfaces whenever I first get on a new pair of skis! How the skis are weighted could also have an effect (lean back, balanced, or lean forward). Surface area and skier weight, maybe?

Basically, I think slope angle would be one of the most variable aspects depending on the other factors.

riverc0il

NewEnglandSkier13;c-35151 wroteIt depends on the snow conditions.

flbski

Roughly 7' or 8' to 1' for a green trail, 10' to 1' for any sort of sliding connection.

DrJeff

Yup, our skis slide on a fine layer of water between our bases and the snowpack. The friction of the base under pressure sliding over the course, crystalline snow structure melts a fine layer of water as we slide over the snow (that’s why on a sunny day, if you catch the light at the proper angle on some groomed snow you’ll see a shiny area of snow over where the skis tracked. That’s just simply that fine water layer frozen up.

Wax gets very important for a skis glide characteristics, as a dry, non wax ladened base is hydrophillic in nature and will increase the coefficient of friction as the ski slides over that fine layer of melted water and decrease it’s gliding efficiency. A waxed ski, courtesy of the hydrophobic wax layer, will repel the water adhesion to the base and almost in essence allow the ski to slide on liquid ball bearings and travel much easier over snow of any pitch.

Keep ’em waxed and you’ll much easier travel fast

DrJeff

Yup, our skis slide on a fine layer of water between our bases and the snowpack. The friction of the base under pressure sliding over the coarse, crystalline snow structure melts a fine layer of water as we slide over the snow (that’s why on a sunny day, if you catch the light at the proper angle on some groomed snow you’ll see a shiny area of snow over where the skis tracked. That’s just simply that fine water layer frozen up.

Keep ’em waxed and you’ll much easier travel fast

skirick

What about wind resistance and the gravitational pull of the moon?

Bill29

NewEnglandSkier13 said It depends on the snow conditions. In the late 60s I taught parttime at Ward Hill, now Ski Ward.It was a perfect place to learn how to ski, especially on the equipment in those days. The slopes were gentle, with nothing to scare newbees. One day we had a storm that left about a foot of mashed potatoes, heavy wet snow. I rode to the top of the t-bar, pointed my skis straight downhill and pushed off. Halfway down the slope heavy, wet snow piled up on my skis and stopped me dead in my tracks.

ciscokid

Height and Weight of skier lol

mapnut

Being technically oriented, I agree with flbski, who actually answered the question.

rickbolger

mapnut;c-35173 wroteBeing technically oriented, I agree with flbski, who actually answered the question.

6-7 degrees if Pythagoras is still correct.

Is that the slope at which friction, snow conditions, curvature of the earth, etc are overcome by slide-ification?

ski_it

flbski;c-35159 wroteRoughly 7' or 8' to 1' for a green trail, 10' to 1' for any sort of sliding connection.

You mean degrees not minutes, right?
Yes, of course there a gazillion variables, but when someone designs a ski hill from scratch there must be some number they work with. I probably should have phrased it much differently

DrJeff

One should figure out the average pitch of the now defunct Juggernaut trail at Killington, especially the top section above the base of South Ridge, AND THEN ADD A DEGREE OR 2 of pitch, and I've think you've got it! ;) :D

mapnut

Being technically oriented, I agree with flbski, who actually answered the question.

mapnut

From Ski_it: "You mean degrees not minutes, right?"
Answering for flbski, that would be horizontal distance to vertical distance - for example, 8 feet (or inches, or miles) horizontally to one foot vertically. I judge slope by lift statistics. For instance, a relatively flat novice lift might be 1200 feet long with 200 feet of vertical (6:1), and a route down which curves away from the liftline could be 8:1 or flatter.