What is the minimum speed for waterskiing based on body mass?

[Willy mc P]

hi,

I'm currently doing an investigation into the physics of waterskiing in my final year at college, and I am investigating bouyancy. Can anyone out there help with this?

I am interested in finding out the lowest speed that one can go on waterski's relative to one's mass, so any equations or info would be appreciated.

cheers

 

[Molydood]

I'm no expert, but I would have thought it had little to do with boutancy as the skies skim the body of water, as opposed to float in it. More to do with forward speed, curvature of the skies and the persons mass IMO.

Way i look at it you have a frontal area of the ski that is interacting with the water coming towards it. This forward force from the water hitting the ski at speed is translated to upward force due to the angle of the ski (ie not flat, not vertical).

Martin

 

[Chi Meson]

Mollydood is correct. The upward force is best examined in terms of impulse vectors. Specifically, the rate at which a specific amount of water collides with the bottom of the skis at a certain velocity is (mv)/t . Notice how the units here are (kg-m)/(s-s) which is the same as a Newton.

The skis have to be at a certain angle, so the upward component of the afore mentioned (mv)/t is what must balance the downward force of gravity. THis is fun because you will have to find the component of a component.

The trick is finding the ratio of m/t. That is, how many kilograms of water contact the bottom surface of the ski in one second when going at a certain speed. One way (in which many unreal assumptions are made for the sake of simplicity) is to multiply surface area of the skis times the velocity. THis will give you "volume of water per second" (check the units). Use water density to convert to mass per second.

Your variables will end up being: surface area, A; velocity, v; angle of attack, theta; and total mass of skier and skis, m.

 

[j-m]

stressed and needing help!

hello all

we have a similar problem as the waterski and are trying to look into the problem experimentally as well as theoretically, but are finding it difficult to actually find ways to measure a lot of the forces etc needed.

people hae suggested using bernoulli's law, but we are having issues grasping the concept

we are also having difficulties understanding what causes the lift force

any thoughts would be greatly appreciated

thanks!

 

[j-m]

Chi Meson your previous post was very helpful but we do not understand what equations you use with the variables?

 

[lpfr]

I'm sorry, but you simply cannot write equations and solve them. This is a problem of hydrodynamics in turbulent flow with almost no symmetry. It is almost "as simple" as to compute the lift of the shuttle.
The only thing you can say are generalities about lift and drag.
In practice water-skis are built my trial and error. It is cheaper and funnier.

 

[rcgldr]

The lowest limit of speed would depend on how big the skis were, and the maximum amount of tension the skier could hang on with, since this would in turn limit the maximum angle of attack.

Normal bouyancy isn't at work here. The skis are riding at the boundary of the air and water, and there's hysteresis effect involved called "planing".

Both powerboats and skis will raise significantly out of the water, reducing drag, and the speed at which this transition occurs is at a faster speed when transitioning from "non-planing" to "planing" as opposed from "planing" to "non-planing". In other words, from a standing start, a certain speed will need to be reached before the skis start "planing", but once they are "planing", the speed can be reduced and they will continue to "plane" until the speed is reduced significantly.

What the limit on the design of the "skis". What if a skier used the equivalent of a twin hull catermaran with tunneling effect to get some lift from the air as well as the water?

Is the skier riding in the wake of the towing boat or off to one side?

The starting point is force equals mass time acceleration. You know the force, since weight would be known, and tension in the tow line can be measured.

As far as collecting data, there are setups where a pole to the side of a boat with a very short or no ski line at all could be used to safely get a profile view of a skier.

One easy calculation though: tension in the tow line (assuming it's horizontal) in pounds times the speed of the skier in mph divided by 375 will give the amount of horsepower consumed in the process.

Getting back to issues, the arbirtrary call for how much water is affected At what point do you consider a pressure differential so small as to be insigificant and not consider the water experiencing this small pressure differential to be part of the total water affected by the skier? Similarly, at what amount of actual water displacment, do you no longer consider the water to be displaced by the skier?

lift of the shuttle

That's easy. When it's orbiting in space, sitting stationary on the ground, or propelled vertically the lift is zero. The other cases are a bit more complicated.

 

[madphysics]

In practice water-skis are built my trial and error. It is cheaper and funnier.

Most skies are made to look good, and to be wide enough to have sufficient surface tension to keep up with the force.

 

[rcgldr]

Most skies are made to look good, and to be wide enough to have sufficient surface tension to keep up with the force.

If people can ski barefooted, then the shape of the ski probably isn't a big factor for a recreational skier. Just keep the front tips curved up so the skier doesn't face plant if the skier drops the tips. From what I understand, unlike snow skis, water skis aren't that senstitive to the weight of the skier. Instead, the towing boat simply adjusts it's speed until the skier is planing, and then any additional speed after that is optional.

 

[madphysics]

Barefoot sking needs much faster speeds to work then regular skiing, and it also has a smaller foot to water contact area.

 

[j-m]

how would we actually measure the lift force of the water ski?

we have the lift force equation as 1/2 * rho * velocity ^2

but we have also seen it as lift = 1/2 * rho * velocity ^ 2 * area * coefficient

but using that what would the coefficient and rho be?

 

[rcgldr]

Those equations for lift force in the air. The coefficient is the coefficient of lift. Water skiing is different, it occurs at the boundary between air and water, so it's a surface phenomenon. There's not a lot of water flow across the top of a ski so most of the "lift" occurs at the bottom of the ski. Here the pressure diffferential is that of the water below versus the air above, a very different situation than that of a aircraft in the air.

Power can be calculated based on tension in a horizontal ski line and the speed of the skier. Power consumed by aerodynamic drag should able to be caclulated via measurement as well (wind tunnel, towing a skier on ice or someother low friction surface). Then the rate of work done on the water can be calculated by subtracting the aerodynamic drag losses from the total power consumed by a skier.

Still even knowing the rate of work done on the water doesn't explain how much water is diplaced forwards, sideways, or downwards (or maybe even backwards), the speed of the flow or how large an a cross sectional area of water is affected.

 

[madphysics]

The speed at which the boat is moving pulls the skier up from the water and in the process the skis press against the water, and because the water is a) thick enough when the skier moves at high speeds to support the skies(and the rider) and b) the surface tension of the water holds up the skies, because at higher speeds, the water is harder.

This like the penny dropping off the empire state building and making a huge dent.

 

[Nuclear on the Rocks]

since the motion involves turbulent flow, it is very difficult to translate into equations unless you use principles of Chaos Theory.
the situation involves viscous drag force of tubulent water + drag force of air

 

[tetis]

If you calculate the force acting when a fluid with density [tex]\rho[/tex] and velocity v flows through a narrowing which transverse section is S, you'll obtain [tex]\rho v^2 S [/tex]. But if you think of the ski just like a narrowing which transverse section is [tex]S = tan(\alpha) A [/tex] and you think of the fluid like it is moving with respect to a firm ski you'll obtain, for the vertical component, [tex]\rho v^2 A tan^2(\alpha) [/tex].

Aniway in real world the things goes in a different way because the viscosity is relevant in such a way the fluid which velocity changes, when the ski pass over it, isn't the entire section but just only a layer, and all the dynamics isn't stationary but evolving in time.

 

[JaredJames]

tetis, this thread is over 6 years old, with the last post over 3.5 years ago.

 

[Delta2]

This thread is revived every 3-4 years lol.