OK this sounds like a stupid question but someone at the pool told me that my hand needs to be open with my fingers apart when I swim. i normally cup it which is what I thought was the best way. Today I tried my fingers together and my thumb out a little bit from the hand.
The cupped hand seemed best but what do the experts say?
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Um, no, I don't think I am confused, but then if I were I probably wouldn't think I was...
The F in F=ma equation can be elaborate to:
F = the sum of all forces acting on the body
the main components of the sum are
F(p) the propulsive forces exerted by the swimmer
F(d) the drag forces exerted by the water on the body
then F = F(p) + F(d)
noting that F(p) and F(d) will have opposite signs.
if the swimmer has constant velocity then a = 0 and ma=0 so
F(p) + F(d) = 0
or
F(p) = -F(d)
If the swimmer is exerting no propulsive force then
F(d) = ma
and
a = F(d)/m
with drag force F(d) being exerted in the opposite direction of movement the swimmer will decelerate.
All of this is only to say that the original passage was misleading, F=ma does apply but in terms of understanding how much propulsive force you need to apply it is more useful to go to the direct equation for a swimmer moving at constant velocity F(p) = F(d) as swimmer generated acceleration is only involved when pushing off the blocks or wall.
Putting my original question another way, I wonder what the ratio is of drag forces acting on the hand as it moves backward is relative to the force due to the inertia of the water that is being accelerated backward.
My recollection is that the orginal nonsense about lift forces came out of the observation that the hand left the water at a point ahead of where it entered, therefore it couldn't be moving water back to achieve propulsive force and therefore propulsion had to be explained by sculling motions.
As a thought experiment one can image a paddleboat with paddle wheels moving at a constant velocity. The boat can move forward without requiring the paddles to accelerate. The question is how much of the propulsion is due to the paddles accelerating water backward and how much is due to the drag on the paddles as they move backward.
Um, no, I don't think I am confused, but then if I were I probably wouldn't think I was...
The F in F=ma equation can be elaborate to:
F = the sum of all forces acting on the body
the main components of the sum are
F(p) the propulsive forces exerted by the swimmer
F(d) the drag forces exerted by the water on the body
then F = F(p) + F(d)
noting that F(p) and F(d) will have opposite signs.
if the swimmer has constant velocity then a = 0 and ma=0 so
F(p) + F(d) = 0
or
F(p) = -F(d)
If the swimmer is exerting no propulsive force then
F(d) = ma
and
a = F(d)/m
with drag force F(d) being exerted in the opposite direction of movement the swimmer will decelerate.
All of this is only to say that the original passage was misleading, F=ma does apply but in terms of understanding how much propulsive force you need to apply it is more useful to go to the direct equation for a swimmer moving at constant velocity F(p) = F(d) as swimmer generated acceleration is only involved when pushing off the blocks or wall.
Putting my original question another way, I wonder what the ratio is of drag forces acting on the hand as it moves backward is relative to the force due to the inertia of the water that is being accelerated backward.
My recollection is that the orginal nonsense about lift forces came out of the observation that the hand left the water at a point ahead of where it entered, therefore it couldn't be moving water back to achieve propulsive force and therefore propulsion had to be explained by sculling motions.
As a thought experiment one can image a paddleboat with paddle wheels moving at a constant velocity. The boat can move forward without requiring the paddles to accelerate. The question is how much of the propulsion is due to the paddles accelerating water backward and how much is due to the drag on the paddles as they move backward.
