I watched the video on Schoeman's start swimswam.com/.../ and it raised a question I have had for a long time;why jump straight out from the start? Schoeman noted another swimmer who dove slightly up at the start and "stalled out"..In a previous thread Rich Abrahams said a coach told him the same thing about stalling out.The physics of this statement make no sense to me.Horizontal velocity is going to remain fairly constant,vertical velocity will decrease as one goes up and then increase again past the apex. I emailed Brent Rushall and he said to jump straight out or slightly down,but the article he referenced said
" Enter the water steeper rather than flatter (this should reduce the amount of splash (irrelevant water movement)).
Practice diving out as far as possible (maximal horizontal velocity produced primarily by leg drive off the block) before entering the water.
Dive deep so that resistance is reduced and more effective double-leg kicks are executed before surfacing."
To maximize distance(diving as far out as possible) one should angle up about 35-40 degrees(if the top of the block was even with surface of the water it would be 45 degrees(Rob Copeland said 32 degrees in another thread but: en.wikipedia.org/.../Ballistic_trajectory )
No one still does that,but some really good starters used to 1984 Olympic Men's 100m Breaststroke final - Steve Lundquist - YouTube .
When I ask coaches why the start should be straight out instead of angled up I never get an answer other than it has been found to be faster.In researching "found to be faster" I have found very little real confirmation.The best study I found(which I can no longer find the reference for) stated that the most important variable in speed to 15M was clean entry and that the greatest correlation with clean entry was experience.This also means that studies that just compare speed to 15M of different starts need to take experience with the start into account.
When I try the straight out start I have variable success with my entry(as would be expected with a new start.)I am willing to practice to get more consistent if I can get an explanation of why it is faster that makes "physics sense". I have seen too many trends in swimming change to think something is right just because everyone does it.(The first lesson I learned about starting was "ït is not a good start if it doesn't give you a red chest". I have been variously taught to swim freestyle without rolling and to kick out on BR kick so I know common wisdom isn't always wise.)
Parents
Former Member
First, I think Pete and Allen are over simplifying things. If you are ignoring water induced drag, then you are just discussing jumping.
In my most humble opinion of course :angel:
I hadn't looked closely enough at Peter's analysis.If it leads to the conclusion that the 0 degree starter goes further in the air it must be flawed.
It is possible for a 0 degree starter who creates significantly more force when propelling themselves at 0 degrees compared to 30 degree to go further.
I couldn't get Pete's applet to work for me, so I just stole his numbers and plugged them into the projection formulas. Given Pete's numbers, you are correct.
a = angle (0 or 30 degrees)
v = velocity (3.5, 3.8, 4.55 m/s)
t = time
g = gravity (9.81 m/s^2)
y = vertical distance (or height)
x = horizontal distance
sin(30) = .5
cos(30) = .866
sin(0) = 0
cos(0) = 1
y = vt sin(a) - .5gt^2
x = vt cos(a)
The block is .75m tall, so solve for -.75, or when the swimmer hits the water.
30 degrees @ 3.5m/s
solve for t: -4.905t^2 + 1.75t + .75 = 0
t = .608
3.5(.608)(.866) = 1.843m
0 degrees @ 3.8m/s
solve for t: -4.905t^2 + 0t + .75 = 0
t = .391
3.8(.391)(1) = 1.4858m
0 degrees @ 4.55m/s
t = .391
4.55(.391)(1) = 1.77905m
How are you going to calculate the transfer of velocities from air to water given different entry angles?
First, I think Pete and Allen are over simplifying things. If you are ignoring water induced drag, then you are just discussing jumping.
In my most humble opinion of course :angel:
I hadn't looked closely enough at Peter's analysis.If it leads to the conclusion that the 0 degree starter goes further in the air it must be flawed.
It is possible for a 0 degree starter who creates significantly more force when propelling themselves at 0 degrees compared to 30 degree to go further.
I couldn't get Pete's applet to work for me, so I just stole his numbers and plugged them into the projection formulas. Given Pete's numbers, you are correct.
a = angle (0 or 30 degrees)
v = velocity (3.5, 3.8, 4.55 m/s)
t = time
g = gravity (9.81 m/s^2)
y = vertical distance (or height)
x = horizontal distance
sin(30) = .5
cos(30) = .866
sin(0) = 0
cos(0) = 1
y = vt sin(a) - .5gt^2
x = vt cos(a)
The block is .75m tall, so solve for -.75, or when the swimmer hits the water.
30 degrees @ 3.5m/s
solve for t: -4.905t^2 + 1.75t + .75 = 0
t = .608
3.5(.608)(.866) = 1.843m
0 degrees @ 3.8m/s
solve for t: -4.905t^2 + 0t + .75 = 0
t = .391
3.8(.391)(1) = 1.4858m
0 degrees @ 4.55m/s
t = .391
4.55(.391)(1) = 1.77905m
How are you going to calculate the transfer of velocities from air to water given different entry angles?
