Recent content by mikeym

Oh yeh. And here's a link to http://www.getoutofglasgow.com/Guides.htm#Physics"

Thanks for the help! It's much appreciated.

Which gives the conclusion that the maximum fall factor is not 2 as is accepted in the climbing community but f = \frac{d}{L} = \frac{2H+x}{L} = \frac{2H}{L}+\frac{x}{L} And x = \%x L f = \frac{2H}{L}+\frac{\%x L}{L} Where the height H = the length of rope out L f_{max} =...

OK I messed up that conversion of terms across to my values. d \not= 2H d = 2H + x where H is the height above the protection that the fall occurs and x is the distance the stretches and d is the distance fallen. So I can show stewartcs' equation F_{max} = mg +...

That's quite a significant difference to the dynamic of the factors involved. If I rewrite it into my terms d for distance and x for stretch x = \frac{mg + \sqrt{(mg)^2 + 2mgkd}}{k} and F_{max} = mg + \sqrt{(mg)^2 + 2mgA \frac{d}{L}} That puts weight back as the dominant...

Stewartcs, I'm not sure what it is you're doing here. \Delta U_g = mg \Delta y = mg(-2H - d) [SIZE="3"]? From the way I'm interpreting what you've written for your definitions of H and d 2H = d So mg(-2H - d) doesn't make sense? Redbelly98, don't bother with the...

OK the potential problems I'm seeing with this way of calculation are: It takes into account the friction on the carabiner for calculating the force on the pulley but not how the stretch propagates along the rope. It ignores the protection + friction + rope drag on the protection that...

Yes, I've been a fan of evolv shoes since I bought a pair in a sale. :) First off I've got a question about what we've got so far. How did you get the 1/2 in the energy equation Energy = \frac{1}{2} k x^2 [SIZE="5"]? I can't see where I went wrong. Is it something to do with the...

Thanks that's a big help. I might try contacting the manufacturers rather than trying to measure A myself. Hopefully this will allow me to get started with some calculations.

Time to type the original + Time to try my way of getting 15 posts + Time to work out why I'd been locked out of that post + Time to post a new thread complaining about it + Time to work out what was and was not counted as a web address (apparently and thing with www[dot]anthing is enough) +...

Does a = F/m = 2 gd / x not show g the gravitational pull which will result in a maximum stretch x? If so I'm not so interested in the whole space man climbing thing :). F = k x = 2 mgd / x thanks but I'm still not sure how to use this because I don't know what x will be. So k could be...

I have to admit that I am probably a one poster as I have one very specific question, but I think that my post is perfectly valid, in which case as I said the rules are quite obstructive. (I've spend a couple of hours messing about with something that should have taken a couple of minutes.)

Hi, I'm a climber and I'm trying to understand the physics involved in climbing and falling. In particular I'm trying to understand the factors that are used within the climbing community to describe it and to unite this with my (high school) physics understanding. I also would like to be...

OK, but how do I report it to you?

In the meantime it would be nice to be able to make my posts.