I believe that when you scale displacement from say a 1:1 big boat that you would scale by the ^3 of the reciprocal of the scale factor, correct?

Example: 12,000 pounds big boat turned into a 1:12th scale rc boat would be d=12000/12^3 -> d=6.944 lbs

Now when scaling an RC boat, down by say half, it is purely just on the scale factor?

Example: 50in Marblehead and I want to scale it down to a 25 inch boat. Would the displacement simply be divided in half or divided by 8 (2^3)? Thank you very much for your help.

Divided by eight unless you want an interesting looking submarine!

Pete

The cube of the scale function remains without regard to what size object you are scaleing. That is an evil reality in some cases.

Disappointing example: a 50/800 might weigh 8 pounds. Now you wish to scale the 50" boat down to 25". The all up displacement of the smaller boat would need to be one pound. Not likely to be able to do that. At maybe two pounds, or any other weight exceeding one pound, the little boat is overloaded and can not run on its’ lines. What to do? You must re-do the lines of the smaller boat so that it can carry the weight. That usually involves adding a bunch more rocker or fattening the mid sections. Trouble with that is that it is no longer the same boat.

If you are detrmined to do this, including line modification, you will find this number useful. (.03611)… That is the number of pounds of water in one cubic inch. Make a realistic estimate of the total weight of the small model. Say 4 pounds including ballast. Divide 4 by 0.03611 to get the number of cubic inches of flotation that you’ll need. 4/.03611 = 110.8 cubic inches. So far so good. Now divide that number by the designed waterline length. Say…20 inches. 110.8/20=5.53 square inches. We are getting there. Now if the boat had the same section all the way from the bow to the transom things would be simple. We’d need a section whose immersed area is 5.53 in^2. But you are going to whittle away some of the front end and some of the back end. You are going to whittle away something on the order of 45% of the total volume of the an object that has the constant shape of the mid section. Now we have a nuber to work with. The number is called prismatic co-efficient. You whittled away 45% and that left 55% of the original shape. The Cp is then .55. If the ends of the boat are real fine, say like a kayak the Cp will be smaller. Conversely if the ends are fat then you have not whittled away so much and the Cp will be higher. Conventional boats like an IOM or similar can use th ball park figure 0.55…

Whoooeeey that was a bunch of explaining.I hope it makes some kind of sense. We need an average section with nderwater area of 5.53 square inches. But that is the average area. We need to know the real area of the mid section not the average. Divide 5.53 by the Cp. 5.53/.55=10.05 square inches. Now you know how big to make the mid section. This is a method that establishes the mid section dimensions approximately, not absolutely. That is because we had to make a SWAG estimate of the prismatic coefficient. You can design as you go along. This method will give you a place to start.

Have fun and keep an eraser handy.

Luckily, I am not trying to do the example, it was merely that. But I had just been thinking about, after I posted it made much more sense to me once I ran a few more instances through my head. Thank you for the help.

messabout, thanks for the back-of-the-envelope discusssion/example.

Only one comment - it’s MUCH easier if you use metric. 1 gram of water takes up 1 cubic centimetre = 1 millilitre, so all the volume/weight conversions are simply a matter of moving the decimal point.

:devil3: