OK, more modelling, of skiers starting on the slope or in the air. I think this includes all 3 suggested scenarios. What I am not sure is what the posters are trying to achieve with these scenarios, but what the heck.
The air skier is teleported/unhooked from a helicopter/whatever at the same time as the slope skier is allowed to start.
Model: 45 degree slope, 4m on the horizontal for a slope length of 5.66m. g=9.82 (because that is what it is where I live). The slope runs from (0,4) to (4,0)
Same for each model:
Skier 1 starts at (0,4) at rest.
He accelerates at g*sin45 along the slope, and reaches (4,0) after 1.28 seconds.
Skier 2 starts at (2,4)
He drops 2m accelerating at g, hitting the slope after 0.64 seconds. He converts a proportion of his impact velocity equivalent to sin45 into velocity along the slope, continues accelerating at g*sin45, and reaches (4,0) after 1.06 seconds. He wins, but did start ahead.
Skier 2 starts at (0,6)
He drops 2m as before, accelerates down the slope and finishes at 1.36 seconds. He loses.
Skier 2 starts at (2cos45,4+2sin45) (to maintain the 2m radius)
He drops 22sin45, converts his velocity and accelerates to finish in 1.28 seconds. The same time when rounded (but actually about 3/1000 slower).
So how the finish works out depends very much on where you start the skiers (unsurprisingly). Given that both are starting from static, this is not very realistic: skiers don’t in my experience pop into existence above slopes. Also, I am assuming that the landings are perfect. YMMV. For a dynamic air solution, I think my post above comes fairly close to a good answer (once you allow for my typo with the axes). Watching some skiing last night the airtime predicted looked credible along with difference in finish times.
No, it’s not about 3/1000 slower; that’s rounding errors. It’s the exact same time.
Skier 1 takes exactly sqrt(16/g) seconds to reach the finishing point. More generally, after t seconds, skier 1 has moved a distance of g/sqrt(2) * t^2/2 along the slope. Furthermore, after t seconds, skier 1 has acquired a velocity of g/sqrt(2) * t parallel to the slope.
Skier 2 in situation 3 starts at position (2 * cos(45), 4 + 2 * sin(45)) = (sqrt(2), 4 + sqrt(2)), and takes exactly sqrt(2 * 2 * sqrt(2)/g) seconds to fall the distance of 2 * sqrt(2) onto the slope at point (sqrt(2), 4 - sqrt(2)), which is at a distance of 2 from skier 1’s starting point (0, 4). That is, skier 2 lands in exactly the same place after sqrt(2 * 2 * sqrt(2)/g) seconds as skier 1 would also be.
Skier 2 also has just acquired a downwards velocity of g * sqrt(2 * 2 * sqrt(2)/g) at this point, which breaks down into components parallel and perpendicular to the slope of magnitude g * sqrt(2 * 2 * sqrt(2)/g) / sqrt(2) each. The normal force of the slope destroys the perpendicular component, leaving only the velocity parallel to the slope of magnitude g * sqrt(2 * 2 * sqrt(2)/g) / sqrt(2) = g/sqrt(2) * sqrt(2 * 2 * sqrt(2)/g). That is, after sqrt(2 * 2 * sqrt(2)/g) seconds, skier 2 also has the exact same velocity as skier 1 would have.
So after sqrt(2 * 2 * sqrt(2)/g) seconds, both skier 1 and skier 2 are on the slope at the point (sqrt(2), 4 - sqrt(2)), with a velocity of magnitude g/sqrt(2) * sqrt(2 * 2 * sqrt(2)/g) pointed along the slope. Accordingly, they continue in the exact same fashion, and both end up the finishing point (4, 0) at the exact same time, sqrt(16/g) seconds after the start.
>No, it’s not about 3/1000 slower; that’s rounding errors. It’s the exact same time.
OK
I just set up a spreadsheet to run the figures so I could play around with different scenarios, didn’t actually run the formulae down by hand. Kudos.
Also apologies for being a thread hog. Unless I come up with anything world changing, or to answer specific points/correct my own stupid errors, I’ll hold off for a while.
I think my slope change model comes closest so far to giving an answer, but it is still very simplistic. For example, the centre of mass of the skier will still follow (or try to follow) the ballistic pathway at the slope change - that’s what gives you that funny feeling when you go over a humped bridge for example. Thinking about skiers again, when they come to a bump, they often tuck lower, to bring the centre of mass down, so that when it enters its ballistic phase it still comes in close to the simpler line of my model. They can maintain contact with the slope after the change by extending their bodies.
So during the immediate pre-change period upto the end of the ballistic phase, the ground reaction forces are going to be somewhat reduced (effectively the g force will have a little anti-g force opposing it, so the normal and accelerative forces will be reduced).
Net result: it gets closer and closer timewise. However, the courses are generally a lot longer than 60m, and the skiers can/will catch ‘air’ many times throughout the course, and a lot of little differences will start to add up. If you additionally input potential lot of control when landing, increased air drag during flight, possible loss of the best line due to flying over it, etc, then staying on the snow is likely better. But it is very complicated, and certainly too much for a few lines on a spreadsheet. Anyone got Matlab and a day to spare?
Conclusion: air looks bad, for a whole bunch of complicated reasons. Anyone else got a strange sense of déjà vu?
