Dives: Once and for all, how much faster do they make you?

Well, it depends on the quality of your dive and the speed of your stroke. If you have a dive that does not have much air-time, then the difference between the dive and the push-off is minimized. In fact, because of the altitude of the starting point of the dive, if air-time is minimized, then the starting altitude and proximal entry into the water will combine for a deep dive, and, in that case, the push-off may very well be faster, due to the additional time it takes one to surface from a curvilinear path (as occurs in a deep dive). However, if you have a great launch, then your speed through the air is markedly better than at the surface of the water or under water, due to the reduced resistance one experiences while traveling through gases vs. liquids. Because of this greatly reduced resistance through the air vs. water, you enter the water, on a good launch, with just about as much velocity, from what I've seen, as when you push off the wall. So, if dive entry and push-off match up for velocity, then the distance traveled in a given amount of time from each of those points will also be roughly proportional. Thus, subtracting out dive entry and the push-off itself from each, the applicable measure for determining the time delta of dives vs. push-offs seems to be to compare the distance one travels in the air and time one takes during launch vs. the amount of time it takes to swim that same distance. If it takes one full second for the swimmer pushing off to make up for the additional distance through the air the diving swimmer gained by the time they both surfaced, then that's the amount of time benefit. Sooooo, assuming two equally matched Phelps, Phelps does 100m in 47.5, or did once upon a time. That's .475 seconds for every meter, which is roughly .16 seconds per foot. If the distance traveled in the air is, what, nine feet-ish?, then we're at a 1.425 second benefit to the diver Phelps vs. the push-off Phelps, minus whatever fraction of time it takes to travel that distance in the air. At least, that's my best guess. --Sean

Well, it depends on the quality of your dive and the speed of your stroke. If you have a dive that does not have much air-time, then the difference between the dive and the push-off is minimized. In fact, because of the altitude of the starting point of the dive, if air-time is minimized, then the starting altitude and proximal entry into the water will combine for a deep dive, and, in that case, the push-off may very well be faster, due to the additional time it takes one to surface from a curvilinear path (as occurs in a deep dive). However, if you have a great launch, then your speed through the air is markedly better than at the surface of the water or under water, due to the reduced resistance one experiences while traveling through gases vs. liquids. Because of this greatly reduced resistance through the air vs. water, you enter the water, on a good launch, with just about as much velocity, from what I've seen, as when you push off the wall. So, if dive entry and push-off match up for velocity, then the distance traveled in a given amount of time from each of those points will also be roughly proportional. Thus, subtracting out dive entry and the push-off itself from each, the applicable measure for determining the time delta of dives vs. push-offs seems to be to compare the distance one travels in the air and time one takes during launch vs. the amount of time it takes to swim that same distance. If it takes one full second for the swimmer pushing off to make up for the additional distance through the air the diving swimmer gained by the time they both surfaced, then that's the amount of time benefit. Sooooo, assuming two equally matched Phelps, Phelps does 100m in 47.5, or did once upon a time. That's .475 seconds for every meter, which is roughly .16 seconds per foot. If the distance traveled in the air is, what, nine feet-ish?, then we're at a 1.425 second benefit to the diver Phelps vs. the push-off Phelps, minus whatever fraction of time it takes to travel that distance in the air. At least, that's my best guess. --Sean