SummitPost.org

See this trip report:

http://www.summitpost.org/trip-report/1 ... ntain.html

After the fact someone mentioned on a forum that Breckenridge had 130 mph gust that day, but I have no idea what the wind speed was on North Star.

Moni wrote:Wind speed estimate for those of us who don't pack calculators in our climbing packs. Fred and I aborted a climb a few weeks ago, because the wind was knocking us off balance while boulder hopping. You could do the lean into the wind thing as in the first frame of this thread (and, yes, it was funny when I fell over once when the wind died back). We figure it was a steady 45 mph with gusts of 50+ or so.

Thanks. useful link to estimate the wind speed at the top of mountain with clues given before ascending.

While i suppose this is a fairly theoretical discussion, I would think that there's a limit to how much you can physically lean forward and still have your feet sticking to the ground. E.g. at 65 degrees or so you would have to be balancing on your toes or the front end of your boots, the area of which is pretty low, not to mention your centre of gravity is no-way near them. So I imagine you would slip and fall on your face / get blown away (so to speak). Any thoughts?

Baarb wrote:While i suppose this is a fairly theoretical discussion, I would think that there's a limit to how much you can physically lean forward and still have your feet sticking to the ground. E.g. at 65 degrees or so you would have to be balancing on your toes or the front end of your boots, the area of which is pretty low, not to mention your centre of gravity is no-way near them. So I imagine you would slip and fall on your face / get blown away (so to speak). Any thoughts?

Yes. I saw another post mentioning this earlier, but I can't find it now.

At some point, the combination of wind speed and angle of lean would create significant lift on the person, and the person's feet would have to be secured to the ground in order to stay in place at that and higher wind speeds. And, of course, well before the person was lifted from the ground, the force of the wind on the person, coupled with the reduced feet-to-ground friction caused by the upward lift, would make the person slide backwards.

Baarb wrote:While i suppose this is a fairly theoretical discussion,

Yes, the model ignores some effects that may become quite important for large values of alpha. Besides, it ignores stability. If I'm leaning forward 10 degrees and the wind suddenly dies, I move one foot forward and regain balance. Not so easy if I'm leaning 45 degrees into the wind. Even such a simplified model, though, is useful in understanding what goes on.

Baarb wrote: I would think that there's a limit to how much you can physically lean forward and still have your feet sticking to the ground.

Indeed there is--even according to the simplified model. If the angle and the coefficient of friction are such that tan(alpha) > mu, then your feet no longer stick to the ground.

Baarb wrote: E.g. at 65 degrees or so you would have to be balancing on your toes or the front end of your boots, the area of which is pretty low,

Friction is a complex phenomenon, but in first approximation Coulomb's Law applies, according to which the friction force is independent of contact area.

Baarb wrote: not to mention your centre of gravity is no-way near them.

True, but that is accounted for in the equation for v that hansw posted at the beginning of this thread.

Day Hiker wrote:Yes. I saw another post mentioning this earlier, but I can't find it now.

That was my post, which I edited, because I found an error in my computation that invalidated the conclusions.

Day Hiker wrote:At some point, the combination of wind speed and angle of lean would create significant lift on the person, and the person's feet would have to be secured to the ground in order to stay in place at that and higher wind speeds.

The lift is proportional to cos(alpha)*sin(alpha), which we learned in high school to equal sin(2*alpha)/2. Therefore, the lift is maximum for alpha=pi/4. The cos(alpha) factor accounts for the reduction in effective cross section, while the sin(alpha) factor accounts for the fact that the force of the wind is perpendicular to the climber (or cylinder, because that is how the climber is modeled in this case); as the climber rotates, a greater proportion of that force is upward.

There are obviously 2 variables in the equation- "What wind speed will keep you from falling down?" doesn't provide enough information (Like being given a word problem saying 2 trains start moving toward each other at time t; they are x miles apart; when do they pass each other? If you don't know how fast they're going, you can't answer the question). Obviously, if you are leaning forward at 5 degrees, much less wind will be required to keep you from falling down than if you're leaning at 30 degrees...

Diggler wrote:2 trains start moving toward each other at time t; they are x miles apart; when do they pass each other?

They pass each other at time t+x/(v1+v2). Maybe I'm missing your point, but hansw gave v as a function of alpha as answer. What's wrong with that?

brenta wrote:

Diggler wrote:2 trains start moving toward each other at time t; they are x miles apart; when do they pass each other?

They pass each other at time t+x/(v1+v2). Maybe I'm missing your point, but hansw gave v as a function of alpha as answer. What's wrong with that?

I guess that a simple equation is the answer (i.e. one doesn't get an answer unless (at least) 2 values are provided).

nickels wrote:If we are assuming that the wind always comes straight on, then one would have to do an alpha sweep of the human body (presumably in a wind tunnel or by simulation) to calculate the equilibrium speed, methinks.

The wind tunnel experiment would provide more accurate data, but--by itself--no insight into what's at work. The simulation model presupposes the kind of insight provided by the back-of-the-envelope analysis (and more).

nickels wrote:I wouldn't think that analytical solutions would be of much use except maybe for the straight up case, where someone has already calculated a terminal speed??

You cannot withstand any significant wind if you insist in standing up straight. The analytical solution tells you that, without giving you precise numbers.

nickels wrote: I guess kind of the same thing that Diggler is saying above....

Not sure. He explicitly refers to the dependence of the speed on the angle. You are much more concerned, as far as I can see, with uncertainties in drag coefficient and mass and area distribution. Once again, I may be missing your point.

Diggler wrote:I guess that a simple equation is the answer (i.e. one doesn't get an answer unless (at least) 2 values are provided).

That could be argued in more than one way. Why waste your time solving the problem for just one set of inputs? The whole function tells you much more than one of its points. It's also a matter of milieu. In functional programming circles--and not only there--the kind of distinction you make between functions and data is regarded as a bad thing.

But all this assumes the wind remains constant in strength - which it very rarely does. Last time I tried this on a summit the wind suddenly died down and I ended up on my hands and knees on the ground!

Geeks!

How about this one: Solve for x & y!

I'm curious to know the conversion factors & algebra behind my recent stress. Is there a calculation for how long it will take me, before I decide to go riot in the streets?

What is the ratio/conversion factor/algebra/wind speed/calculus/ for the following:
1) the amount of money I've lost in the stock market this past year
2) the amount of money the fat cats will walk away with
2) the amount of money I have to bail out their fat asses
4) the amount of time it will take me to recover the money

Given: It took me 22 years to save the money in balanced, conservative funds.

i love threads like this.

some of us seriously have way to much time on our hands.

All of this ignores altitude...the wind at 20,000 feet has significantly force than the same speed at sea level.