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 hat. 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 draf 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 one 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 200 ft = 600m]

W = Fdrag xD = 1/2pv2CAd

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

W = 13,440,000 Joules.

W = 3.7 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 7.4 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 7.4 kWH = 111 cents = $1.11.After all the approximating we’ve done I feel like it could be anywhere from $1.11 to $20.

This seems very low. 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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