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Thread: Explanation of DIN?
01-15-2010, 02:03 PM #1Registered User
- Join Date
- Jan 2010
Explanation of DIN?
I've cruised this forum, and understand pretty well how it works, and I apologize for my first post not including nekkid pics of my girlfriend, but I'm at work right now and that would be decidedly NSFW. I'm hoping putting this in the correct forum atleast counts for something.
ANYway, I'm doing some research here at work and I've been charged with the task of explaining how ski bindings work. I know about forward pressure, DIN that sort of thing, but what I'm curious to figure out is what exactly happens when you tighten the DIN? I know it compresses a spring, but how does that then translate to making it harder to release? How does the binding then release when certain force is exerted upon it?
EDIT: also any pictures/diagrams of the internal workings of the ski binding would be awesome.
Any and all help appreciated. I apologize for being a JONG, but we all have to start somewhere
01-15-2010, 02:35 PM #2
I think you should take a pair of bindings apart and figure it out for yourself.
Plus, google is helpful.
01-15-2010, 02:51 PM #3
when you get home, put up the pics. then you'll be told.
01-15-2010, 02:58 PM #4
Let me google that for you, JONG!
All your answers can be found here.
Last edited by JayPowHound; 01-19-2010 at 05:55 PM.
01-15-2010, 03:16 PM #5
For a JONG, at least he is being very courteous and has made a promise of pics, so I'll play nice. This is what DIN is, it\\\'s an organization that sets standards
They set a standard that puts the amount of torque (typically measured in foot pounds, inch pounds or newtons) required to release a ski binding into a simple scale of 1-whatever.
A spring sets the tension required for release, IIRC compressing the spring increases the amount of pressure (and consequently torque) required to release said binding.
Basically, the internals of a binding (and consequently your boot) can slide back and forth, the spring inhibits this. the amount it inhibits this motion is the DIN scale.
/binding jong acting like he knows what the hell he is talking about
01-15-2010, 03:16 PM #6
Good first post.
It's evident that you took a little time to at least figure out this community a little before posting your question. As oftpiste mentions however, you have to give before you get...
Therefore, nekkid picks for the win, otherwise hit Google and don't come back.
01-15-2010, 10:54 PM #7
dumpy, minor correction, metric units of torque are Newton-meters."It need not be fun to be fun." - Big Steve
01-16-2010, 12:16 AM #8
Didn't Include Nekkid.
01-16-2010, 09:19 PM #9
I'm actually curious to know what the ft/lbs of torque are that are represented by a din number and what is the difference between each number? Is it the same difference in torque up and down the scale or is it logarithmic? Does half a DIN number make a difference for a type III+ skier?
inquiring minds want to know.No longer stuck.
01-16-2010, 10:02 PM #10"It need not be fun to be fun." - Big Steve
01-17-2010, 09:28 AM #11
lol, that's what those numbers are on the chart. duh.
but how do I know how much I'm changing when I go from 8.5 to 9 with a 315 BSL? Is that a significant change?
it seems like the tolerance is logarithmic.
I guess I'm asking about spring weight or whatever that's called. I have no idea what I'm asking.No longer stuck.
01-17-2010, 10:00 AM #12"Fakers are Maggots" - T. Hall, 2011
only a fake Rasta could make a claim like that
01-19-2010, 02:38 PM #13
now tell me what it is in ft/lbs and in english tell me what kind of bearing that has on my ACLs.No longer stuck.
01-19-2010, 05:48 PM #14
Interesting... so if I'm reading this right, if I were to decrease my BSL by about 15 (new boots with smaller shells) I should increase my DIN setting (by about just under .5) in order to achieve the same release torque value my knees are used to?
I know it's basically insignificant... but just sayin