any math/science people?

[edt]

I dont think there is any significant energy stored in the lines at all

So it does not matter what weight you are, and the "spring" effect in the lines can be neglected regarding jump height IMO.

Huh! I've read a couple of "How to's" in magazines recently that talk about 'loading up the lines' and using the spring to get height ... perpetuating old wives' tales?

Well I think both Peter and I know the physics and math required to do the calculation . . .

But. It's windy tomorrow, and the day after, so I'm not doing it any time soon.

[alexrider]

I dont think there is any significant energy stored in the lines at all

So it does not matter what weight you are, and the "spring" effect in the lines can be neglected regarding jump height IMO.

Huh! I've read a couple of "How to's" in magazines recently that talk about 'loading up the lines' and using the spring to get height ... perpetuating old wives' tales?

Loading the lines is a energy conversion technique as is the bending of vaulter's pole, and is a ramp converting horizontal momentum to vertical momentum, which is turned into height.
If any energy is stored in the lines, it will be at the expense of the rider's speed.
So what Peter Frank meant was: "I dont' think there is any extra energy stored in the lines at all ".
Don Monnot is the one who started to ask about static jumping; so his quiz deserves a reply.
"I know I can jump while just standing on the beach".
Yes. But you will lose ground and gain backward velocity.

[Kamikuza]

Can't remember the exact words, but the impression they left on me was that you were using the lines as a bungy to BOING up into the air ... of course, my memory might be faulty

What about ... wind shear as the kite goes up, you get higher winds and so jump EVEN higher?

[joriws]

I'd say load&pop in terms of lines is not true. I mean storing more energy into tension (length change to have a spring) of the lines.

Usually lines are pre-stretched/tensioned to 300/500kg they are more pre-stretched than you can generate force by load by turning upwind.

Also lines are not fixed to a solid object. You kite will change it's shape a bit (I guess more on C-kite than bridled-bow-hybrid -ones) and you'll pull/force your kite upwind by riding 1 meter upwind, so you pull the kite. Similar when you run on the ground pulling kite lines.

So my cents are with load'n'pop that you accelerate kite's forward flying speed a bit by loading and take some 'shape slack' away by making fabric more tight/stiff and reactive as kite is more 'rigid'.

Also about the physics, I really do not agree with Peter's energy theorem because kinetic energy is speed at surface perpendicular (90 degrees) to potential energy. So for kinetic energy velocity vector you need to redirect that upwards. Velocity change is done by acceleration, acceleration is force and force is generated by kite's lift which has relation to airspeed, angle of attack and lift/drag curves). Why cannot kite generate more power to have more energy upwards taken from wind than you originally had when cruising? Because acceleration would allow it. 3 options for energy, you have less, equal or more energy than cruising

As for hang glider stuff. You should not think these lift-issues on your ground/water speed. Wing has it's own relative airspeed which generates the lift. If wing is flying it's speed can never be equal to apparent wind so that apparent wind is "still" because wing will not fly.

But on ground coordinate you can have 10m/s apparent wind and wing's forward speed 10m/s, then you are still on ground coordinates. I've done hang gliding landings on normal forward ground speed, to a stand-still and even flying backwards relative to ground. But wing/lift is there.

Also if true wind (20m/s) is blowing and you are flying upwind (15m/s forward airspeed), relative to ground you move backwards (5m/s), but still your airspeed is 15m/s. And still wing's aerodynamical efficiency and sink rate -rules are valid so you glide downwards (as you fly forward 15m/s) by our sink rate determined by your wing's characteristics.

Wings have their speed polar, lift/drag ratios etc. as well and so do the kites. This is why kites differ. Lift co-efficient min-max -scale = depower ratio. Lift/drag is glide ratio.

Also take a though of wing loading and sink rate when you estimate hang times. You need to optimize your lift device sink rate and initial height to gain hang time. If you can jump 10m high, 1m/s sink rate will allow you to glide for 10s, 2m/s for 5s. (simplified a lot). Wing loading is mass per wing area, think of comparing small umbrella and a parachute when you jump off from a cliff.

The truth is somewhere there...

[edt]

Also about the physics, I really do not agree with Peter's energy theorem because kinetic energy is speed at surface perpendicular (90 degrees) to potential energy. So for kinetic energy velocity vector you need to redirect that upwards. Velocity change is done by acceleration, acceleration is force and force is generated by kite's lift which has relation to airspeed, angle of attack and lift/drag curves). Why cannot kite generate more power to have more energy upwards taken from wind than you originally had when cruising? Because acceleration would allow it. 3 options for energy, you have less, equal or more energy than cruising

Tried several times to understand that paragraph. I failed. Absolutely zero idea what you are trying to say.

[Pablo]

Also about the physics, I really do not agree with Peter's energy theorem because kinetic energy is speed at surface perpendicular (90 degrees) to potential energy. So for kinetic energy velocity vector you need to redirect that upwards. Velocity change is done by acceleration, acceleration is force and force is generated by kite's lift which has relation to airspeed, angle of attack and lift/drag curves). Why cannot kite generate more power to have more energy upwards taken from wind than you originally had when cruising? Because acceleration would allow it. 3 options for energy, you have less, equal or more energy than cruising

ahhh, i c now, i had to read this 3 times to get what u mean.
partially agree. in this case the kinetic energy u have is kinda useless cos you have no ramp or anything to convert it into height (potential energy)
the only thing you have is the kite, and it doesnt care which is the source of it wind or how fast you're moving relative to the ground. the only important thing is the apparent wind it feels.

on the other hand when you load the kite, suddenly it's "feeling" it's deeper into the wind window, so it will pull more till it reaches the end of the new wind window again (based on your new direction). so from this point of view, the faster you were riding, the further it'll get into the window and so more pull. but still this is true to generate force paralel to the ground.

so at the end i think it's just a matter of not letting the kite win (go in the wind direction, and therefore kill the pull). This last sec edging changes your path, increasing the apparent wind speed (and so increasing the lift) plus loading the kite as i explained above.
it also lets you use your legs better.

at least that's how i see it.

[joriws]

Also about the physics, I really do not agree with Peter's energy theorem because kinetic energy is speed at surface perpendicular (90 degrees) to potential energy. So for kinetic energy velocity vector you need to redirect that upwards. Velocity change is done by acceleration, acceleration is force and force is generated by kite's lift which has relation to airspeed, angle of attack and lift/drag curves). Why cannot kite generate more power to have more energy upwards taken from wind than you originally had when cruising? Because acceleration would allow it. 3 options for energy, you have less, equal or more energy than cruising

Tried several times to understand that paragraph. I failed. Absolutely zero idea what you are trying to say.

Maybe I need to open it more as I was writing it in a hurry And probably it is needless to say that I am not a native English-speaker.

- Kinetic energy= 0.5 * m * v^2 is speed energy of kiteboarder usually 90 degrees to gravity field.
- Potential energy = m*g*h is energy stored in gravity field, g is acceleration constant and height is h. How high you are above zero level (sea level) is the potential energy.

Now Peter's suggested energy conservation law is kinetic energy = potential energy. But they are 90 degree's off in direction. Potential energy is towards zenith (gravity field) and kiteboarders cruising speed is towards 3 or 9 o'clock (at same gravity level 0) so kinetic energy and it's velocity vector is towards 3 or 9. Naturally air resistance (friction) is omitted which constantly shortens the velocity vector as friction constantly decelerates you.

How we could turn that kinetic energy into potential energy. We need to change kinetic energy velocity vector upwards. Speed/velocity is altered with force and kite provides that force to accelerate kiteboarder into a new direction. How kite can generate force depends on it's area, airspeed, lift co-efficient (on some Angle of attack) and air density. When kite is redirecting, why it cannot also increase speed (increase vector lenght) at the same time for even more kinetic energy.

So calculating maximum jumping height from equation 0.5 * v^2 = g * h ***> h = 0.5 * v^2 / g is not necessarily true if you can add kite energy (v * v or v^2) at pop with good kite control. So at jump point you can have less energy, equal energy or even more energy compared to kinetic energy when you we gaining speed for the jump.

After pop when you leave water you are a paraglider with a big swing (20-30m). This swing can be used to gain even more lift/energy from the kite (megalooping) by using your weight to pull the kite. Kite lift vector component in paraller to g's direction will add height if lift force is greater than your combined mass. And then helicopter looping down to have higher lift from your kite and to limit speed.

[edt]

When kite is redirecting, why it cannot also increase speed (increase vector lenght) at the same time for even more kinetic energy.

Still doesn't make sense. So a heavier kite will lift you more because it has more kinetic energy? <puzzled emoticon>

One thing I realized reading some of Peter's posts is that our maximum velocity is pretty constant. That is it can blow 40 knots or 14 knots we have about the same board speed. I think this must also mean that we keep about the same board speed with slack lines or with loaded lines. Sure with loaded lines it will be a little less, but I bet you can store energy in the lines. So maybe there is something to this whole "loading up the lines" thing.

As for changing kinetic energy to potential energy (height), in Peter's example we just assume that the kite acts like a pole vault, perfectly converting the energy, it doesn't matter what direction the kinetic energy vector is, down up, sideways, just assume you convert it to height using the kite to change the direction of motion. This will give maximum height, assuming the lines can't store energy. The velocity should be measured not from water speed but in terms of air speed.

[oldkiter]

Can someone graph "wind speed" - mph or knots - versus jump height - feet or meters??

Obviously need to keep many variables the same - rider weight (maybe use 75kg/165#) - board size (maybe use 135 x 40 TT and/or 6' directional) - etc, etc, etc.

So graph would probably be theoretical maximum heights - assuming flat water.

Thanks!!

[edt]

Can someone graph "wind speed" - mph or knots - versus jump height - feet or meters??

Obviously need to keep many variables the same - rider weight (maybe use 75kg/165#) - board size (maybe use 135 x 40 TT and/or 6' directional) - etc, etc, etc.

So graph would probably be theoretical maximum heights - assuming flat water.

Thanks!!

if you look at the equations board size, rider weight does not matter, basically they "cancel out" when looking at the equations. Given that we ride right around 20 mph, no matter what the wind speed is, usually around 90 degrees to the wind, v = sqrt(20^2 + wind^2), pe = mgh = ke = .5 mv^2, height of jump = .5 (20^2 + wind^2) / g, g is 22 mph per second, 3600 seconds per hour, and we want the final result in feet instead of miles so multiply by 5280, chart (.5(20^2 + wind^2) / (22 * 3600) ) * 5280, you get these results

mph jump height in feet
10 17
15 21
20 27
25 34
30 43
35 54
40 67
45 81



.This are maximums, not taking into account friction of the air, water, nor technique nor the realistic board speeds. But I think they give you an idea of this method.

You can open up excel or something if you want a graph of those numbers.