I work out pretty regularly and I go pretty heavy when I do, but the pump I get from flying fixed bridal far surpasses anything I have accomplished in the gym. So it got me wondering what are the FT LBS that are transfered through the handles to your arms while flying a FB kite (no strop). I know there can be a lot of variables, like aspect ratio, position in the wind window, etc...

So for arguments sake, let's say : what are the FT LBS that are transfered through the handles to your arms while flying a low aspect 2m FB kite in 25mph winds....? What about G's ? What is the potential G force that we are exposed to in such conditions....?
****edit****. since position of the kite is a variable, let say dead center of the powerzone.
Discuss...

(MIke Dobbs, I expect you to be all over this...)

It's going to be variable because the amount of pull differs depending on where the kite is in the wind window.

You also have to factor that speed = power. The kite will develop more pull when you "work it." On a moderately windy day I can casually fly the Rage 2.5 or 3.5, and the pull will be manageable. If I work it agressively, I will have to lean back and dig my heel in to hold my ground. It sort of the effect you get when you really crank a Powerwball up.

If you were really, really ambitious, I expect you could hook up some type of strain gauge to the handles/leaders to measure actual pull.

As far as "G's" I think you have to be in motion for that to apply. The are G-Force apps for the iPhone and I assume the Android. Standalone unit's have been around for years in the aftermarket car scene. I expect people will do double takes if you mount one to your buggy and they realize what it is.

ATB,
Sam

ATB,
Sam

My last session out, I noticed when I was sining the kite with the strop running through the hook in my spreader bar, that it got hot to the touch. The pull from the kite coupled with the friction of the strop rubbing the hook under pressure actually heated the strop up enough to be hot to the touch when it contacted my forearm. This was my 5.5 Reactor II so it can make some power.

G forces can't really be much higher than your ability to hold down the power. I've pushed it a time or 2 like at the sod farm in the thick grass coming out of a turn with a montana 9.5 and get on the power, I've been stood up on the footpegs then settle back into the seat when the "Gs" settled back down.

You could always use a scale to measure the pulling force of the kite. What I mean is the type of scale that fishermen/hunters use to hang their catch on. you could hook one end up to a spreader bar and the other on a strop, then when the kite pulls it will measure how many pounds/kilos of force that is applied to the strop. Simple and cheap way to sooth the curiosity.

I don't think G's are a factor. You're only pulling 1 G unless you are in transit. Now, if we're talking G's in acceleration or G's in a transitional turn then that's a different story.

And in motion the you have resistence that is independent of G forces. Example: hard edging against the kite while being propelled forward. The G's are in the forward motion not the resistence against the wing.

If I'm wrong here someone let me know.

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Originally posted by Feyd I don't think G's are a factor. You're only pulling 1 G unless you are in transit. Now, if we're talking G's in acceleration or G's in a transitional turn then that's a different story. And in motion the you have resistence that is independent of G forces. Example: hard edging against the kite while being propelled forward. The G's are in the forward motion not the resistence against the wing. If I'm wrong here someone let me know.

I kinda had the same thought but wasn't sure...

i dont know anything about this but i can tell you the max will be the breaking strain of your lines.

Gs are for accelleration or deceleration only and can occur from turning, starting, stopping, or being lifted upward. All the other stuff is just force.

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Originally posted by BeamerBob Gs are for accelleration or deceleration only and can occur from turning, starting, stopping, or being lifted upward. All the other stuff is just force.

Well now, the more I think about it, what about lateral G's? Such as when you are cornering in a car ? The opposing pull against the kite must be similar.... No ? :dunno:

When cornering the car, you are accelerating sideways due to the force of the car on you, so the kite isn't accelerating you sideways if you are resisting it and just going forward at a steady direction and speed. but it can accelerate you away from your straight line creating G force.

Hey everyone- great discussion so far... only just got home.

Right off the bat Cheezy, if you really want to know the answer take Cheddarhead’s advice and hook up a strain gauge. That being said, let's get some terminology out of the way.

What you are interested in is FORCE=(mass)(acceleration) and POWER = FORCE*DISTANCE/TIME. POWER is a measure of how much force you can apply in a given amount of time.

The amount of force required to move you a given speed is of course dependent upon the coefficients of friction between the tires and the surface (which we don't know) so trying to calculate it would be pretty iffy. With a direct measurement (as Cheddarhead suggests) it would be relatively simple to calculate work done.

Now, G-Force refers to acceleration. Acceleration is a change in Velocity, so if there's no change in velocity there's no acceleration and no G-Force. One G refers to a change in acceleration equivalent to the force exerted on us by gravity which is approx. -9.8m/s/s or -32ft/s/s Notice the units, distance per second per second. In other words, change in speed per second, or acceleration (negative in the case of gravity)
So, let's say you wanted to pull 1G with your kite from a standstill. If you weight about 90kg (mass with your buggy), and want to accelerate forward at 9.8m/s/s, this would require a FORCE = (mass)(acceleration) = (90kg)(9.8m/s/s) = 882 Newtons which is approx. 650 lb-ft. Wow- that seems like a lot to me! :shocked2: :shocked2:

I don't think I could hold on to more than about 260lbs (based on my deadlift), so if my numbers are right, pulling a G without a harness isn't in my near future!

Hope this helps some… Next time we’re out it might be fun to try to nail down some of these numbers a little more exactly- could be fun!

**EDIT** In these calculations we are talking unbalanced force, i.e. the force the kite is producing beyond the force required to overcome friction (and air resistance which is negligible)

**EDIT** thanks to GeoKite, I left distance out of my power equation

Thanks Mike for the more highly formal explanation. You might be stronger than you think. Slalom and jump water skiers have been strain gauged to create around 1000 lbs force when crossing the wake at peak pull. This is only for a fraction of a second but still a high number. They aren't harnessed in either of course, straight to the hands.

BeamerBob you make a good point- I'm sure that for a brief instant I could hold down more force- and now that I think about it, grip isn't really the limiting factor in my deadlift.

All the more reason to try and get some hard data to pursue this further : )

This is an awesome question. I may have to write up a kite loop transition question for my differential equations final this semester. If it makes the cut I will be sure to forward it along.

Cheeze:

Dunno 'bout no math n stuff
But I think it's more simple than physics...

Just the constant resistance of holding the handles and pulling them in n out at steady, various rates (instead of jerk n lift) should give you a more intense kind of work out.

Plus, it's way more fun and challenging than bars or a machine, so the time and effort isn't noticed as much (until you're sore the next day!)

Of course, when there is no wind, you have to find an adequate substitute to hold your interest for that kind of intense, workout motion



& for those w/ Hooj Ballz:

Quote:

Originally posted by MikeDobbs What you are interested in is FORCE=(mass)(acceleration) and POWER = FORCE/TIME. POWER is a measure of how much force you can apply in a given amount of time.

Power is energy/time. Power is how fast energy is used.

Example: Walking up stairs requires less power than running up stairs. A car with more horsepower can expend more energy per time than a car with less horsepower. A 100W light bulb uses more energy per time than a 20W light bulb

(I'll be teaching this in two months to my students)

Quote:

Originally posted by geokite Quote: Originally posted by MikeDobbs What you are interested in is FORCE=(mass)(acceleration) and POWER = FORCE/TIME. POWER is a measure of how much force you can apply in a given amount of time. Power is energy/time. Power is how fast energy is used. Example: Walking up stairs requires less power than running up stairs. A car with more horsepower can expend more energy per time than a car with less horsepower. A 100W light bulb uses more energy per time than a 20W light bulb (I'll be teaching this in two months to my students)

Thanks Geokite, yes- slight misstep on my part. Power is (Force*Distance)/time, or work/time.

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Originally posted by 43patrick This is an awesome question. I may have to write up a kite loop transition question for my differential equations final this semester. If it makes the cut I will be sure to forward it along.

This would be an easier one to calculate theoretically because we wouldn't have any friction with a surface to take into account (assuming you are referring to an aerial transition). If we look at some video of kiteboarders transitioning in air, we may be able to determine their velocity. If we know the velocity before, the velocity after, and the time to make that change, we could calculate the average acceleration over that time (assuming acceleration to be constant), then back into an average force at the harness.

The amount of force exerted by the kite is equal to the the force from the acceleration of the mass + the force from any stretch in the lines + force to to resistance of movement, (air,kinetic friction etc..)... this is a spring mass problem with the lines being the springs with a very high spring constant. mx''+Bx'+kx=f(t). The problem is that in real life these are all three dimensional vector and B, k and f(t) not the easiest functions to identify, We could probably come up with reasonable estimates a for a simplified example.

I could be wrong but I thought this was the equation for G-Force:

G-force = Guinea Pigs + spy gear.

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Originally posted by van I could be wrong but I thought this was the equation for G-Force: G-force = Guinea Pigs + spy gear.

LOL too funny

Just back home from the berkshires and saw this. I have nothing to add. Just wanted to let you know I'm here in spirit.
Oh and I don't work out either. But I do work hard.

the amount of force would probably be easiest to find with an aeronautics equation for lift and drag instead of a physics equation even though it would still technically be a physics equation