My guess as to top speed follows. Thunderchine is, I believe, a nominally 55lb motor, so let's assume that. You can plug in other figures later if needed.

First off, convert to newtons. 55lb ≈ 244N

Newton tells us that Force = Mass x Acceleration

Assuming zero drag, we could simply plug that in and get the acceleration, and thus the speed at any given distance travelled. This might give us a useful upper bound, but it's going to be "optimistic". So, we need to look at drag and friction.

First off, aerodynamic drag.

Let's assume the kick has a drag coefficient similar to that of a bicycle or a skier, or around 1.0. This handily gives us a drag force of ½ρv²A, where ρ is the volumetric mass density of air, and A is the frontal area of the kick. Assuming a temperature between ±10C on the lake, we can estimate ρ at around 1.3. At a guess, the frontal area of the kick is around 0.5m²

So, at any given time, the aerodynamic drag force is going to be of the order of 0.325v², where v is the velocity in m/s relative to the medium (we'll assume still air for simplicity)

Next up, frictional losses.

Steel on ice has a very low coefficient of friction, somewhere between 0.004 (for special speed skates) and 0.1. We'll assume 0.01, as a relatively reasonable figure scientifically plucked from my arse (and based on the fact the ice is unikely to be mirror-smooth)

The kick itself will be running horizontally, so the normal force is simply the weight of the kick + rider * g, or 1177N, giving a frictional force of 11N or so regardless of speed.

We'll ignore the static friction by assuming that the thrust of thunderchine is enough to overcome this. If it's not, you're fucked, but you'll melt a nice pool of water while you're trying to get going.

So, what does this give us?

Well, the driving force, Ft, at all times will be F(thunderchine) - F(friction), or 233N

The aerodynamic drag force, Fd, at a given instant, will be 0.325v²

The acceleration, A, at any given moment is therefore (Ft - Fd) / 120, or (233 - 0.325v²) / 120

Velocity, v, at any given time t is the integral from 0 to t of A.

Distance travelled, x, at any given time, is the integral from 0 to t of v.

Rather than buggering about integrating all this twice and then back solving for x = 2000, which I was never any good at even when I could remember how to do it, let's write some code.

Code: Select all

```
#!/usr/bin/ruby
x = 0.0
v = 0.0
t = 0.0
a = 0.0
dt = 0.001
# calculate until we get to the magic 2000m mark
while x < 2000
t += dt
a = (233.0 - (0.325 * v * v)) / 120.0
x += v * dt
v += a * dt
end
puts "t = #{t}s, x = #{x}m, v = #{v}m/s, a = #{a}m/s^2"
```

Which gives me this result

Code: Select all

```
t = 84.2540000000382s, x = 2000.00155627203m, v = 26.7751553335021m/s, a = 3.82845668307634e-05m/s^2
```

Obviously, the reult varies depending on the choice of dt, but the orders of magnitude are the same even if we go down to the scale of seconds rather than thousandths.

27m/s is 97km/h or about 60mph.

[edit] It occurs to me that the rated 55lbf is "sans ejectors", and that their performance will vary depending on forward speed. Any estimate as to how much they will add, m?