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I'm working on a hoverboard project and am having trouble finding out the maximum weight these fans could lift.

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  • $\begingroup$ How is a question about building a hoverboard on-topic here? $\endgroup$ – Ralph J Mar 18 '19 at 13:52
  • $\begingroup$ Weren't there a bunch of hoverboard questions around 18mo ago?? $\endgroup$ – acpilot Mar 19 '19 at 3:29
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You have volumetric flow rate and diameter, so you have the $\Delta v$ ($v = A \dot V$) and you have the mass flow rate ($\dot m = \varrho \dot V$) and that gives you the momentum per unit of time a.k.a. force ($\dot p = \Delta v\cdot\dot m = F$).

Run the numbers yourself, but don't get high hopes—the result is tiny (of course; you need a couple of metres diameter to lift a human with useful efficiency).

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  • $\begingroup$ Of course, the amount of force is much higher if the hovercraft is operated in ground effect, in which case the amount of force is simply $F=\Delta p \times A$ with $\Delta p$ the pressure difference across the fans (probably a function of leakage flow at the edge of the hovercraft) and $A$ the surface area of the hovercraft. $\endgroup$ – Sanchises Mar 18 '19 at 9:37
  • $\begingroup$ @Sanchises, but can you actually derive $\Delta p$ from the known parameters? Also I wouldn't really agree with the qualification “much”. It will probably be several times, but given how tiny the force actually is, it does not make it much more practical. $\endgroup$ – Jan Hudec Mar 18 '19 at 18:50
  • $\begingroup$ No, I don't suppose you can, that's entirely dependent on the fan characteristics. It makes a large difference for a hovercraft if these fans could deliver this volumetric flow at hundreds of kPa or a few Pa. I guess my main point is that for a hovercraft, you don't want thrust but $\Delta p$. Then again I'm not what a hoverboard is. $\endgroup$ – Sanchises Mar 19 '19 at 7:24
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Taking the conclusion of Hudec's answer:

$ \Delta v\cdot\dot m = F$

First of all, let's use decent SI units...

3 x 160 cu ft/min = 0.22653 cubic meters/sec. As they pass through those fan apertures (total 0.07444 sq. meters) in one second, the airspeed of that mass is 0.22653/0.07444 = 3.0431 m/s. If the change of speed is from zero, then $ \Delta v\ =$ 3.0431 m/s

Now, 160 cu ft of air = 0.07551 cu. meters = 0.0928 kg (density of air = 1,23 kg/m3)

Inserting numbers in Hudec's expression: 3.0431 m/s x 0.0928 kg/s = 0.2822 N

A very tiny force indeed...

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  • $\begingroup$ You used the flow from one fan and the area of all three, so your velocity is wrong. $\endgroup$ – amI Mar 18 '19 at 10:12
  • $\begingroup$ @aml You're right. Correction introduced... $\endgroup$ – xxavier Mar 18 '19 at 12:01
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(Jan Hudec's answer is correct but probably hard to understand.)
You want Force and you know flow (cfm =$ft^3/minute$) and area ($\pi r^2$). Force is mass times its acceleration. Mass is weight/gravitational acceleration, and the acceleration of that mass is its change in velocity over time, so $F=(W/g)*(\Delta v/t)$. The starting velocity of the air is zero, and each fan gives it a velocity of flow/area. The weight of air moved in a given time is flow*density (sea level air weighs about 0.076 lb/cu.ft.).
Make sure that you use the same units throughout (for length, time and weight) and multiply by the 3 fans when done. This force is the total thrust and therefore the maximum weight that you could fly freely (you will find that it is very small).
If you are making a 'hovercraft' with a skirt then you are just reducing friction rather than flying (the pressure of the trapped air comes into play).

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  • $\begingroup$ please consider using MathJax markup for mathematical symbols and equations. also, a linebreak here and there helps with readability. $\endgroup$ – Federico Mar 18 '19 at 12:17
  • $\begingroup$ @Federico - thank you -- is there a tutorial I should see? $\endgroup$ – amI Mar 18 '19 at 17:32
  • $\begingroup$ if you ever used LaTeX, the syntax is basically the same. Otherwise, you can have a look at the official website: mathjax.org/#docs $\endgroup$ – Federico Mar 18 '19 at 17:36

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