Damn Math

So, I am doing my research and here is something I came across.
The important part is this:

One thirtieth scale is much too small for meaningful observations. For example there is the weight scaling problem. Use the scale factor cubed...30 x 30 x 30 = 27000..... That is the divisor to use and the weight of the full sized boat is used as the dividend....Example: suppose the full sized boat is to weigh 2700 pounds. 2700/27000= 0.1.....your model will need to displace 0.1 pounds to simulate the 2700 pound boat. That is not a practical weight for the model. Now consider using something on the order of one eighth scale for the model. Scale factor 8 cubed = 512...and your 2700 pound big boat is simulated by 2700/512 = 5.27 pounds. which is a more realistic number. If you are thinking of a small planing boat that might weigh 800 pounds then the scale weight problem is even more serious.....800/27000 = 0.0296 pounds. If you are working in the SI system then kilogram weights work just as well. Make the model bigger. You will get much better results.

Of course they are talking about a larger boat and so on.  But the math should be the same, right?
I displaced enough to hold 8.146 lbs. If I cube that, I get 540.547 lbs. Pretty good. 

If I use the above formula I figure 1/4 scale (so 4 x 4 x 4 = 64) and multiply that by 8.146 to get 521.344 lbs. Also pretty good.

But are they correct? Both numbers are close enough that I wouldn't have a problem, but what if this is not the right path? The overall  All of this could be off.  I know that this boat could easily hold 500 lbs. But will it hold that with the same waterline? I like the way she sits right now: it's elegant and stable, but I want to know what it will do at full size. 

Well, I did the only good thing to do when you need help: I asked for help. The Meers-Cat is confirming that this is 'pretty correct,' by which I mean 'close enough.'

Dave and I are in negotiations for space in his garage. I guess I should start looking for materials...