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.)
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.
As i understand Peter's analysis he is calculating position from the wall at the time the 30 deg starter enters the pool. Because the 30 deg starter is in the air for all of this time, but the flat starters are already in the water, the flat starters poistion is advancing underwater, albeit at an assumed veloctiy 25% less than there Vx at entry.
As a non physics person, I have nothing to offer EXCEPT
I can beat James Adams off the blocks.
I use a straight forward track start and no competitor beats me to the 15 meter mark.
So I'm not changing.
:)
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:
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 Peter = 1.78m - same]
How are you going to calculate the transfer of velocities from air to water given different entry angles?
He simplified it as follows:
Let's assume that the flat divers lose 25% of their speed in the water for the extra .323 seconds that the 30 degree diver is still in the air. The "weaker" 3.8 m/s swimmer would travel and extra 0.92m in that time while the "stronger" 4.55 m/s swimmer would travel and extra 1.10m.
At entry for the angled dive: here are the stats:
0 deg, 4.55m/s start - 3.88m from the wall
0 deg, 3.8 m/s start - 3.41m from the wall
30 deg, 3.5 m/s start - 3.02m from the wall
Woohoo :banana:.
After that they can answer the age old question of "how fast can you go in a pool filled with diet coke wearing a tech suit made of mentos.":bolt:
Wait ... I think I just got an idea for a new tech suit
He simplified it as follows:
I am not arguing with your arithmetic, rather I think the error must be in his velocity assumptions.The 30 degree velocity reduction is because some of the speed is going into the vertical vector so the horizontal speed will always be less going up,yet we know from the ballistic diagrams that the horizontal distance is always greater,with the maximum distance at 45 degrees.This applies even when the starting point is higher than the landing point.
The 30 degree velocity reduction is because some of the speed is going into the vertical vector so the horizontal speed will always be less going up,yet we know from the ballistic diagrams that the horizontal distance is always greater,with the maximum distance at 45 degrees.This applies even when the starting point is higher than the landing point.
Not quite. The reduction in velocity is due to the vertical component of the force that the legs provide acting against gravity. You can't consider projectile motion until after you have left the blocks and no other forces are being applied to the swimmer.
I'd play with the projectile java applet a little bit. The maximum distance is 45 degrees ONLY when there is no difference between the starting height and ending height. If your starting height is higher than your ending height, the maximum distance will be less than 45 degrees. I can prove this with the equations but playing with the applet would probably be easier to visualize.
Also, ballistics makes the assumption that the exit velocity of a gun barrel is the same no matter what the angle. For a relatively small mass accelerated by a relatively large force, this is a valid enough assumption for engineering work. Consider however the case of a canon that fires a cannonball with the force of 1g. If the barrel is pointed straight up, the cannonball will never exit the barrel. If the barrel is pointed horizontally, the cannonball will exit so long as it can overcome the friction forces inside the barrel.
How are you going to calculate the transfer of velocities from air to water given different entry angles?
I didn't want to go there. As pointed out, I made a very crude assumption that you would lose 25% of your horizontal velocity in something like .3 seconds. I don't know if that is valid. It's been a while since I've looked at fluid dynamics. I'll see if there is an easy way to estimate this but I'd guess that and empirical approach would be much easier.
As a non physics person, I have nothing to offer EXCEPT
I can beat James Adams off the blocks.
I use a straight forward track start and no competitor beats me to the 15 meter mark.
So I'm not changing.
:)
AHA!! A challenge yet again!! Indianapolis. You'll have to enter the 1650 with the same entry time as me so we can get seeded together (only chance of men/women together), and then we'll race it to the 15 meter mark AND compare reaction times off the blocks. I do start the 1650 the same as my 50/100, just don't normally sprint to the 15, but in this case I can make an exception.
Now, can we get you to enter the 1650? :D
AHA!! A challenge yet again!! Indianapolis. You'll have to enter the 1650 with the same entry time as me so we can get seeded together (only chance of men/women together), and then we'll race it to the 15 meter mark AND compare reaction times off the blocks. I do start the 1650 the same as my 50/100, just don't normally sprint to the 15, but in this case I can make an exception.
Now, can we get you to enter the 1650? :D
What?! Waste a nationals entry on the 1650? I will stick with the much harder to place in sprint events. :D
I had one reaction time in Omaha that was 6.0.
I was speaking of my own competitors (I'm sure there are young fast big dudes who can beat me), though I am up for the 15 meter challenge! Non-free.
