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Old     (santa)      Join Date: Jun 2004       10-13-2015, 2:46 PM Reply   
To all you physics and ballistics buffs out there: am I right in thinking that jump height is the only factor that will affect air time?

I mean, we all want more time in the air, which gives us more time for bigger tricks. So if I'm right, then it wouldn't matter whether you are landing barely W2W or way out there in the flats. As long as the maximum height (apex?) of the jump is the same, you would get the exact same air time.

Can anyone with ballistics or physics knowledge confirm this?
Old     (Orange)      Join Date: Jun 2012       10-13-2015, 3:51 PM Reply   
If the water were perfectly flat, then the height of your jump would indeed be the only determinant of air time. The water isn't flat, though, so there could be small differences in air time based on where you land because the vertical distance you drop from your apex is not the same.

As an example, if you jumped to a max height of 6 feet and landed at the same height you left (like a perfect w2w jump), your hang time would be roughly 1.2 seconds (0.6 up, 0.6 down). If you landed past the wake in the trough and that trough were 2 feet lower than the top of the wake, you'd travel 6 feet up and 8 feet down and it would take about 1.3 seconds - the same 0.6 up and 0.7 down. Technically you'd have a little more time to do your trick, but small fractions of seconds.
Old     (King12)      Join Date: Jul 2012       10-13-2015, 6:49 PM Reply   
Any changes in your wind resistance do to differing movements should technically have an effect on hang time.
Old     (ilikebeaverandboats)      Join Date: Jul 2007       10-13-2015, 8:08 PM Reply   
I like that you are roping in ballistics with this question. You silly goose.
Lets assume "Air Time" is the total time in the air (t).

t = 2*V *sin(a)/g

Where...
V = initial velocity at "launch" in the "launch" direction
a = angle from the horizontal to the "launch" direction (sin(a) is the "y component" of the height)
g = gravity (constant)

Neglecting drag, wind direction, coriolis accel, rope stretch, your mom...

Your total time in the air is going to increase as your velocity in the launch direction increases and the angle of your launch direction increases. Here is some data, plot it if you would like. I assumed V at launch to be 18 mph. Notice that at an angle of 90 (straight up) you have longest time in air, go the highest, but you did not go wake to wake...
a V V t g ( a is degrees, first V is mph, second is ft/s, t is seconds g is gravity ft/s/s)
0 18 26.4 0 32.2
10 18 26.4 0.284739869
20 18 26.4 0.560828061
30 18 26.4 0.819875776
40 18 26.4 1.054011981
50 18 26.4 1.256122565
60 18 26.4 1.420066501
70 18 26.4 1.540862434
80 18 26.4 1.614840042
90 18 26.4 1.639751553
100 18 26.4 1.614840042
110 18 26.4 1.540862434

Imagine how factors like wake shape (as a function of speed too), speed a rider can hit a wake and not crumple, board shape, yada yada yada can influence things like speed and the angle.

Research "Time of Flight" if you want to do some reading.
https://en.wikipedia.org/wiki/Projectile_motion
http://formulas.tutorvista.com/physi...n-formula.html
http://www.intmath.com/trigonometric...-amplitude.php

Last edited by ilikebeaverandboats; 10-13-2015 at 8:11 PM. Reason: I should have just attached a screenshot of my xcel spreadsheet, but im that lazy....
Old     (King12)      Join Date: Jul 2012       10-13-2015, 8:34 PM Reply   
^ good info. But wouldn't hat 90 degree angle cause a higher peak height? I was thinking about this and wondering if you had to keep peak height the same velocity would have to be less on that jump thus reducing air time?
Old     (ilikebeaverandboats)      Join Date: Jul 2007       10-13-2015, 10:03 PM Reply   
Quote:
Originally Posted by King12 View Post
^ good info. But wouldn't hat 90 degree angle cause a higher peak height? I was thinking about this and wondering if you had to keep peak height the same velocity would have to be less on that jump thus reducing air time?
My main disclaimer, I don't claim to know much of anything... and I hit my head a lot.... I don't even know what I don't know!

Not sure what you mean good sir. "Peak Height" ?

At 90 degrees, you would be getting the "highest" since your velocity (18mph in my example) is entirely in the y direction. A less than formal (*waits for math guy to attack engineer*) way of looking at the previously posted equation is that the sin(a) determines how "much" of the horizontal velocity component is "converted" into a vertical velocity component. At sin(90)=1 ALL of the horizontal velocity goes up! at sin(0) there is no vertical component (No airtime for you sir!) What do you think happens at 45 deg?

The 90 degree angle is not realistic in our example (especially while maintaining the 18mph), but its nice to see the extremes of an equation (at least for me, it helps show what is going on) and I figure it would be a nice aid for everyone else. Notice that the time frame of around 1.5 seconds is in the right ballpark

Maybe I could have done some nice plots in MATLAB that showed different trajectories for different sets of parameters..would plots make anyone other than myself happy? I LOVE PLOTS.

But they would only represent an ideal world full of nice pointy vectors where no one knows what entropy is, and nothing is turbulent....
Old     (King12)      Join Date: Jul 2012       10-14-2015, 5:36 AM Reply   
Try to explain what I was saying with this graph


These are all with the same initial velocity, just varying the jump angle.

So, OP states in jumps of the same height* which I was taking to mean the riders peak/max height during the jump. In the case that you reduced the jump angle if you kept the same velocity as say a 45 degree jump, a 75 degree angle would mean the rider goes higher... So that doesn't qualify for his scenario.

To fix this, you have to reduce initial velocity until max height under parabola is the exact same as the 45 degree jump.

So it's just a dinky little jump instead of really high or really long. Wouldn't that negate any additional airtime you would get from baying the angle but keeping the same initial velocity?

Disclaimer for me, I'm a biologist not even an engineer, much less a physicist lol. This is fun though

* I jacked this graph real quick from u of Alaska, btw so credit to them

** yes, I love plots too

Last edited by King12; 10-14-2015 at 5:39 AM.
Old     (biggator)      Join Date: Jul 2010       10-14-2015, 6:39 AM Reply   
If the initial velocity is higher, a lower angle can still reach the same max height as a lower velocity/higher angle and should account for greater air time.

The graph above show different angles at the same velocity - now adjust velocity and you'll get the same max height, but farther to the right on the graph.

Make sense?
Old     (fly135)      Join Date: Jun 2004       10-14-2015, 6:53 AM Reply   
A big wake definitely affects air time. Jump height is just a result of all the other factors. But yeah I believe max jump height does correlate with max air time.
Old     (King12)      Join Date: Jul 2012       10-14-2015, 7:14 AM Reply   
Quote:
Originally Posted by biggator View Post
If the initial velocity is higher, a lower angle can still reach the same max height as a lower velocity/higher angle and should account for greater air time.

The graph above show different angles at the same velocity - now adjust velocity and you'll get the same max height, but farther to the right on the graph.

Make sense?

That's what I thought. This is what I said except I used reducing velocity on the greater angle and you used increasing velocity on the lesser angle. But it's the same idea

Last edited by King12; 10-14-2015 at 7:19 AM.
Old     (skiboarder)      Join Date: Oct 2006       10-14-2015, 7:42 AM Reply   
I don't think you can discount the rope. if you clear the tower height, the boat is actually pulling you down so it wouldn't make a smooth arch. I'm not smart enough to begin to start the math.
Old     (jws2)      Join Date: Apr 2015       10-14-2015, 9:14 AM Reply   
the is all very cute however the variables are endless...... example the angle of the turn in relation to the boat is going to vary the rider speed, if the boat is doing 20 mph (just to keep it simple) and he (or she) makes a 45 degree turn (just to keep it simple) then his speed of approach would be more like 40 mph.... (he would travel 2ft for ever 1ft of boat travel) so that would change every thing. The rope would also act like a catapult depending on tension.... I am sure there are other factors I have missed.
Old     (Orange)      Join Date: Jun 2012       10-14-2015, 9:42 AM Reply   
All these answers are saying the same thing, though holding different things constant. The vertical height of your jump is the only thing that determines hang time, though there are an infinite combinations of launch speeds and launch angles that will produce the same height.

Applying it to wakeboarding, if you're trying to maximize your hang time in order to do more tricks, the ONLY thing you need to focus on is getting more height (not distance) on your jump. You can do this in either of two ways, or a combination of both: You can focus on technique to increase your launch angle and get the most height you can out of whatever speed you're traveling, or you can increase the speed at which you hit the wake. Which is right for you really depends where you are today. You'll put your body less at risk if you can improve your technique and pop higher using the same speed you are today, but if your technique is already pretty pure you may not have any choice but to increase your speed (by adding boat speed, taking a more aggressive edge toward the wake, or taking a longer approach with the same edge).

I personally find my technique suffers greatly when I try to use brute force and increase my speed into the wake, so sometimes my jump can actually get worse when I increase my boat speed or get too aggressive with my edge. Sometimes I actually end up with a higher jump by backing off my edge and really focusing on technique. My orthopedic surgeon, however, needs a new Porsche and is trying to get me to go faster.
Old     (ILikebooty)      Join Date: Mar 2015       10-19-2015, 10:46 AM Reply   
I'm sorry but all of you guys are wrong. You're leave out multiple contributing factors that are crucial for determining air time while wakeboarding. Trust me, I just graduated high school, and took an AP stats class... So I know what I'm talking about. Try to keep up. The factors that contribute to air time are: boat speed, line length, hair length (in), and f***s given. The formula to determine this is:

Airtime= (Boat Speed X Line length) + (Hair Length) - (f***s given)

For example, If you ride 28 mph and 90 feet back, have gnarley locks of flow, and couldn't care less. Your formula would be

Airtime= (28 X 90) + (14) - 0 = 2,534

Now, due to the law of wake punctuation (established by Darin Shapiro in 1995), you must replace the comma in your final answer with a period. This turns the large number you get into a more reasonable decimal. So your final answer for the above example is 2.534 seconds of airtime.
So to answer OP's question:
Contributer to most air time = Randall Harris
Old     (Jmaxymek)      Join Date: Feb 2012       10-19-2015, 12:13 PM Reply   
Quote:
Originally Posted by ILikebooty View Post
I'm sorry but all of you guys are wrong. You're leave out multiple contributing factors that are crucial for determining air time while wakeboarding. Trust me, I just graduated high school, and took an AP stats class... So I know what I'm talking about. Try to keep up. The factors that contribute to air time are: boat speed, line length, hair length (in), and f***s given. The formula to determine this is:

Airtime= (Boat Speed X Line length) + (Hair Length) - (f***s given)

For example, If you ride 28 mph and 90 feet back, have gnarley locks of flow, and couldn't care less. Your formula would be

Airtime= (28 X 90) + (14) - 0 = 2,534

Now, due to the law of wake punctuation (established by Darin Shapiro in 1995), you must replace the comma in your final answer with a period. This turns the large number you get into a more reasonable decimal. So your final answer for the above example is 2.534 seconds of airtime.
So to answer OP's question:
Contributer to most air time = Randall Harris
Interesting logic there booty hahaha
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