[Update] I talked to some sources familiar with the Wavepool and I was low on my estimate on the surface area of the foil. I’ve updated the price below.

For now, only a select few have been able to ride the wave pool. Sure, the wave is perfect. But how much would the common person expect to pay to ride it?

One factor to the overall price of the Surf Ranch is going to be electricity. I sought out to find exactly how much energy it takes to make one wave, and then turn that into a dollar amount. I got help from my physics-wiz friend and we did a rough calculation. If anybody see’s anything wrong with the math, feel free to chime in.

Before we get into the physics. If you want a Lemoore Surfing Club shirt. Click Here.

The energy of the wave machine all comes from pushing the hydrofoil through the water. That’s it. The waves themselves are an after effect of the moving hydrofoil paired with a carefully shaped pool bottom.

How much energy does it take to push the hydrofoil the length of the pool? We can think of that energy as the work done to push the hydrofoil the length of the pool.

Work is equal to a constant force (F) multiplied by the distance (D) overwhich the force is applied.

W = F x D

The engine must deliver a force that exactly counters the force of the drag from the water on the hydrofoil. The magnitude of the F-engine must be equal to the F-drag. Thus the work we’re interested in is: W=Fdrag x D

Fluid dynamics are complicated. Simplifying we can assume the draft were dealing with is summed up in this equations.

Fdrag – 1/2pv^2CA. Where P is the density of water. V is the velocity of the foil, C is the drag co-efficient (a number that depends on the shape of the foil) and A is the cross sectional area of the foil.

This is where things get murky. While we know the density of water and the speed of the hydrofoil we don’t know it’s shape and size. I assume its cross sectional area is around two meter squared. I’d assume our value for C is something between .5 and 1. We’ll go with .7.

From this we can get the work i.e the energy in joules. [Note: 18 mph = 8 m/s and 2000 ft = 600m]

W = Fdrag xD = 1/2pv2CAd

W = 1/2 (1000kg/m^2)(8m/s)^2(.7)(2m^2)(600m)

W = 2688000 Joules.

W = 7.4 kWh —-> Energy to push hydrofoil the length of the pool.

The motor won’t be perfectly efficient at turning the electrical energy it’s given into the mechanical energy of the moving hydrofoil. If we assume an efficiency of 50%, we need to supply the motor with 14.8 kWh of energy to make one wave.

Interestingly, 18556 Jackson Ave in Lemoore California was permitted as a single family home and not on an industrial rate plan. This may have changed. Let’s assume the electricity costs are 15cents/kWh.

Thus the cost of one wave is 15 cents / kWh x 14.8 kWH = $14.80.After all the approximating we’ve done I feel like it could be anywhere from $14 to $25.

Let us know if you see anything wrong with our math!

This value does seem a little low, but I appreciate the effort! I don’t have a good sense on the drag coefficient, but the 1m^2 for the projected area of the hydrofoil might be a little low. If the hydrofoil is 2 ft thick by 20 ft long, that comes out to about 3.8 m^2. Getting that sled up to speed looks like it will also take quite a bit of energy. It’s an incredible feat of engineering no matter how much energy it takes!

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Add about a hundred dollars per wave to cover the cost of taxes in California.

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Thanks for calculating. But your conversion from retard units to metric units is not quite right. 200 ft is not 600m. Since I think the pool is actually 600m long, the 200ft might be a typo (2000ft is about 700m), but t just confused me.

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Been thinking on this for a while and pleased to find other technical surf minds that did too. Appreciate the efforts. Seems that return trip of foil would need to be accounted for, although probably not much load. I would guess >$20 per, which is one pricey chairlift ride (esp. if you are prone to louse up the takeoff, like me). 🙂

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I see it says 200ft = 600m. I think you mean approx 60m.

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Hey bud – you were doing fine until you estimated the cross sectional area. I’m watching it right now as we speak – I can see the top section of the foil – it’s clearly 10 m long, height equal to the depth of water which I’d say is at least 1 m deep – therefore it should be more like 10 m^2.

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I’m watching it too! The pool is roughly 700 metres long. Haven’t timed a run of the hydrofoil, but if it’s a minute then the 7kWh above means at least a 400kW motor. Sounds light? If that thing is basically a locomotive then they are more like 2000kW. On that basis the i minute run would be around 30kWh. They are leaving 3 minutes between runs so you get 15 runs an hour so 450kWh per hour.

I’m guessing the capital cost swamps the running cost Though 🙂

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Another way to view this might be to compare it to the cost of running a large diesel-powered boat that distance. I’m no ship’s captain, but somebody around here must have an idea of what a large ship with a similar draw costs to run on a per-mile basis at around that velocity. Note that we aren’t talking about some little ski boat here; it would be one massive yacht to bury itself that deep in the water!

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