#### JSS

• 1350
• Math is math!
##### Re: Lake Minnewanka
« Reply #20 on: September 23, 2020, 01:05:39 AM »
I thought I would do some math on this problem just to see how high those waves would need to be.

Let's assume a flat lake and a distance of 7,000m to a boat 3m high.

That forms a 90 degree triangle with the angle at the camera of 0.024555332573647646 degrees to hit the top of the boat, an extremely shallow angle as expected.

But we can skip all the fancy Pythagorean math and just use a simple ratio.  7000m/3m gives us a ratio of 0.000428571.

So how high would the waves need to be to obscure the boat?

At 7000m they would need to be 7000 x 0.000428571 = 3m high.  Makes sense.
Halfway across the lake at 3500m they would need to be 3500 x 0.000428571 = 1.5m high.

I think it's clear from the video those waves are not 1.5m high. That's 5 feet.

It's hard to judge just how far out those waves are, but considering the zoom of the camera halfway seems reasonable.  The closer they are the shorter they need to be however.  Right in front of the camera they would only need to be perhaps 15cm high, about 6 inches.  Those waves don't look that high though, and the further away you get, the higher they will need to be.  I just don't see waves as a reasonable explanation here.

Those waves don't look anywhere near that high. If the bracket is 10cm high, those waves don't even come close to half that height. Maybe 2 inches at best.