BryanZ Posted February 26, 2009 Report Share Posted February 26, 2009 How do you derive 3 Gs from that picture? just a guess...I think I read somewhere that alpine carving can reach 3g's, and if there's anybody who can reach that, I'd assume it's that guy, in the middle of a race no less. Quote Link to comment Share on other sites More sharing options...
Jack M Posted February 26, 2009 Report Share Posted February 26, 2009 How do you derive 3 Gs from that picture? Let's say 15m board at 60 degrees, 200lb rider... from the math here, that would be turn radius - 7.4m speed - 11.3m/s (about 25mph) = 1570.24 Newtons of centripetal force divide by mass (91kg), divide by g (9.8) = 1.76 Gs. Quote Link to comment Share on other sites More sharing options...
KingCrimson Posted February 26, 2009 Report Share Posted February 26, 2009 well you can't count your own body weight, because when you squat 300 pounds, you dont count yourself. Maybe YOU don't, but some of us have self esteem issues!3 g's seems way high for safe speeds. I'd be willing to bet that I've never broken 1.2, and 2 is probably quite tricky to hit. Jack had something written up about this. EDIT: and there he is. Quote Link to comment Share on other sites More sharing options...
BryanZ Posted February 26, 2009 Report Share Posted February 26, 2009 Let's say 15m board at 60 degrees, 200lb rider...from the math here, that would be turn radius - 7.4m speed - 11.3m/s (about 25mph) = 1570.24 Newtons of centripetal force divide by mass (91kg), divide by g (9.8) = 1.76 Gs. nice man, figured out the g's from the angle.....I don't think we know the turn radius or the speed though. Quote Link to comment Share on other sites More sharing options...
Jack M Posted February 26, 2009 Report Share Posted February 26, 2009 nice man, figured out the g's from the angle.....I don't think we know the turn radius or the speed though. RTFM if you assume a rider weight, sidecut radius and an edge angle, you can work it all out. :p Quote Link to comment Share on other sites More sharing options...
kjl Posted February 26, 2009 Report Share Posted February 26, 2009 If you go the ideal physics/trig route it should be easier than that: #G's = 1/cos(angle) where the angle is how far over your center of gravity is leaned (not how far the board is leaned), where 0 is straight up and 90 is flat on the ground. Standing straight up, 1/cos(0) = 1 G Leaning over 45 degrees, 1/cos(45) = 1.41 Gs Leaning over 60 degrees, 1/cos(60) = 2 Gs 3 Gs would be leaned over ~70 degrees. Although you can get the board almost 90 degrees over, I doubt you can actually get your center of gravity much further over than 70 degrees. but in reality it doesn't really work out that way; both Jack's and my formula's ignore the slope of the run itself (fewer G's at the start of the turn and more at the end), the fact that carving is dyanamic, and you are not really ever at equilibrium so the forces don't actually have to sum to 0, etc.. Quote Link to comment Share on other sites More sharing options...
b0ardski Posted February 27, 2009 Report Share Posted February 27, 2009 Let's say 15m board at 60 degrees, 200lb rider...from the math here, that would be turn radius - 7.4m speed - 11.3m/s (about 25mph) = 1570.24 Newtons of centripetal force divide by mass (91kg), divide by g (9.8) = 1.76 Gs. as i listen to rock lobster after a sweet frozen surf, just push the board thru to the next turn:cool: Quote Link to comment Share on other sites More sharing options...
BryanZ Posted February 27, 2009 Report Share Posted February 27, 2009 If you go the ideal physics/trig route it should be easier than that:#G's = 1/cos(angle) where the angle is how far over your center of gravity is leaned (not how far the board is leaned), where 0 is straight up and 90 is flat on the ground. Standing straight up, 1/cos(0) = 1 G Leaning over 45 degrees, 1/cos(45) = 1.41 Gs Leaning over 60 degrees, 1/cos(60) = 2 Gs 3 Gs would be leaned over ~70 degrees. Although you can get the board almost 90 degrees over, I doubt you can actually get your center of gravity much further over than 70 degrees. but in reality it doesn't really work out that way; both Jack's and my formula's ignore the slope of the run itself (fewer G's at the start of the turn and more at the end), the fact that carving is dyanamic, and you are not really ever at equilibrium so the forces don't actually have to sum to 0, etc.. Looking at all this info, i think it would be safe to say the doing some leg training at the gym is going to greatly improve your carving skills. Most of us probably cheat and straighten our legs out, and that is basically taking away your suspension. Quote Link to comment Share on other sites More sharing options...
Jack M Posted February 27, 2009 Report Share Posted February 27, 2009 I really don't think that's right Ken. Quote Link to comment Share on other sites More sharing options...
Kimo Posted February 27, 2009 Report Share Posted February 27, 2009 An aircraft in coordinated, level turn at a 60deg bank generates 2g. Crank the bank to between 70 and 71 degrees and you get your 3g. This can be loosely applied to carving with the caveat that, while carving, you normally are not executing a level turn. Rather, you are descending and "falling" down the hill thereby relieving some of the G load. Quote Link to comment Share on other sites More sharing options...
Shred Gruumer Posted February 27, 2009 Report Share Posted February 27, 2009 I can touch my Wee Nee on the board when I ride standing up!! And its Cold!! Quote Link to comment Share on other sites More sharing options...
kjl Posted February 27, 2009 Report Share Posted February 27, 2009 I really don't think that's right Ken. It's not right, but it's exactly as right/wrong as your formula. If it is the simplified problem that we were both solving for (carving a perfect circle on perfectly flat snow with no grade in equilibrium - i.e. the carve would be continued forever if there was no friction), then it is exactly right. (Kimo's numbers match mine). You can draw a right triangle with the points being the point of contact between the edge of the board and the snow (P1), the center of mass of the rider (P3), and the point directly below P3 on the snow (P2). If it is perfect equilibrium, the gravity (vector A, which is along the line P2P3) plus the cetrifugal force (vector B along the line P1P2) is perfectly matched by the force given to the rider from the snow (vector C along P1P3). Since it's a right triangle, you can solve for that force, which is just C=sqrt(A^2+B^2), or if you take the angle at P3, you can just do C=1/cos(angle). Note that what you were solving for in your equation was just the force needed to counteract the centrifugal force (so you solved for B, and got 1.76Gs). If you look at exactly how much force is necessary to fight centrifugal force and gravity together, which is what the rider feels, you need to do sqrt(1.76*1.76+1*1) which is 2.02G's, which is just 1% off of my numbers, so we actually agree. The problem with our formulas, as I mentioned before, is that we're not doing, as Kimo calls them, coordinated, level turns - we are descending a slope at a funky angle, which changes things quite a bit, and we are not really at equilibrium - we push and pull our legs and change our angulation and prep for the next turn, all of which changes the g-forces we feel. All that being said, perhaps we get to 3Gs, but I doubt we get much higher. Quote Link to comment Share on other sites More sharing options...
b0ardski Posted February 27, 2009 Report Share Posted February 27, 2009 I touched my knee to the boards20 yrs ago, then I gave up on laces tele & snowboard Quote Link to comment Share on other sites More sharing options...
Dave ESPI Posted February 27, 2009 Report Share Posted February 27, 2009 I firmly tuck my rear knee in behind my front knee... occasionaly THEY touch, but never the board to my knee........ oh lord I PRAY they never do meet....... Quote Link to comment Share on other sites More sharing options...
bobdea Posted February 27, 2009 Report Share Posted February 27, 2009 I firmly tuck my rear knee in behind my front knee... occasionaly THEY touch, but never the board to my knee........ oh lord I PRAY they never do meet....... I think you just explained why you had trouble with that rossi. see Jack's separate the knees article Quote Link to comment Share on other sites More sharing options...
Jack M Posted February 27, 2009 Report Share Posted February 27, 2009 I firmly tuck my rear knee in behind my front knee... that's probably not helping you. Note that what you were solving for in your equation was just the force needed to counteract the centrifugal force (so you solved for B, and got 1.76Gs). If you look at exactly how much force is necessary to fight centrifugal force and gravity together, which is what the rider feels, you need to do sqrt(1.76*1.76+1*1) which is 2.02G's, which is just 1% off of my numbers, so we actually agree. ah yes, you're right, I forgot about the downward component. Doh! And that 1% is due to my rounding. Quote Link to comment Share on other sites More sharing options...
JK moscraciun Posted February 27, 2009 Report Share Posted February 27, 2009 Ok what is the board you are on? Nose like an X5 or is it my imagination?Not saying it is. SG race board...178 or 185 is my guess...I got to see a few in person at the Cypress World Cup...they look amazing, nice clean finish... -Gord is a SG 163 Sl 2010 (see attached) regarding the extreme forward lean of the rear boot as Ken "kjl" stated is an optical illusion resulted from the accumulation of fw flex + binding angle + overall lean toward the slope Quote Link to comment Share on other sites More sharing options...
groovastic Posted February 27, 2009 Report Share Posted February 27, 2009 sorry for OT, but is there any difference between 2009 and 2010 models except for top design? Cheers! Quote Link to comment Share on other sites More sharing options...
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