Bouyoncy of pontoons

Hello, I have built a couple of boats before but I have been really getting into airplanes more lately. I got my pilot license recently and I am currently building an amphibious airplane, which will be ready in the spring. I will be keeping the airplane at the dock at our house. Of course I don't want it in the water all of the time so I am designing a floating lift (water level fluctuates by the hour around here). Basically I will be taking a 2-20' pontoons and building a frame work to bridge them some depth beneath the water. There will be aluminum C channel rails running fore and aft with what is essentially a cantilever style jet ski lift running on these rails. It is a little more complex than that because the rear of the lift must dip down in the loading position and then pick up higher in the storage configuration in order to reduce the lift the wings are making when it is tied down at dock. You will see why this complex design is necessary when you see the landing gear configuration of my airplane along with the sponson, below. In short it is difficult to pull up alongside a dock. Also, see the two pictures of the basic design I have thrown together in solid works.

Now here is where you all come in. I am a couple years out of college as a mechanical engineer (25) and I'm good with the mechanics and design end but buoyancy throws me off a little (never really learned it). When the load is balanced over the pontoons it is a piece of cake to calculate how much water the toons will displace and where the water level will be. However when the lift mechanism is raising the plane the cg will initially be right at that step in the hull (see first pic) which means that I will have a 1100-1400 lbs load located 12-18 inches aft of the rear of the pontoons.

I envision that when I operate the cantilever lift (winch will be mounted to the frame along the front) it will mostly just be sinking the pontoons making the raft do the titanic, not actually lifting the plane. Then once the landing gear (in retracted position shown in picture 2) are above the level of the deck I will engage another winch that will slide the lifting mechanism forward. at some point as the load moves forward the airplane will be lifted fully out of the water. I am trying to calculate what the angle the pontoons will end up at once the gear clears the deck and how far under they would be.

Of course this means that the airplanes hull will come into play some. Also I can add water into the front chamber on the pontoons to act as ballast which likely will be desirable here.

Any ideas on where to start here in order to calculate what angle the pontoons will be at and how far under water the rear will be with a 1400 lb load located 18 inches aft of them? I think initially I should calculate it not taking into account the buoyant force of the airplanes hull.

 

Import your pontoon into MAAT Hydro demo version (free) in the iges format, enter the mass and CG of the pontoon, of the plane and you will make your own calculation very easily.

http://www.sistre-shipdesign-software.com/

 

Just love that plane..a pity it is not VTO.

 

As In Vertical Take Off?

It can take off on land or water in about 250ft. That's good enough for me.

I'll check out that software.

So nobody knows how to do this by hand? It seems like there should be some kind of an established formula for this. I would imagine it would require some integration but being such a basic geometric shape it seems like it would be relatively straight forward. I'm just having a hard time formulating it.

 

No need to have software for naval architecture calculations. With any CAD program, knowing how to use it, it is possible to calculate the submerged volume of the cylinder to the desired angle, for multiple drafts.
Doing it by hand is not complicated but some knowledge of mathematics and a major effort would be required.

 

Mcurcio, can you give us the exact dimensions of the pontoons, and its estimated weight?
It would be even better if you could post a DWG, STP or IGES file of the pontoons.
Cheers

 

The pontoons are a cylinder with a 18in OD, 240 in in length with a wall thickness of .0641" The weight of each pontoon is 113lbs and the weight of the entire raft assembly is about 1000lbs. Again the plane has an empty weight of about 900lbs and a max gross of 1430lbs.

I tried uploading my solid works file of the pontoon but it said invalid file type.

 

As you say, you have all the data from the pontoons, all weights, the 3D model of the pontoons, and a CAD program, very good, which can do the math. What is, therefore your problem?

 

I am looking for a means to calculate how the raft will respond to a load placed at some position on it.

IE. I place a 1000 lb centered between the two pontoons and located 10 ft from the stern (exact middle of craft). Clearly the vessel will now displace an additional 1000lbs causing it to sit level but a few inches deeper in the water. Now I slide that load back so it is still centered between the two pontoons but now it is inline with the rear of the pontoons, Question is at what angle will the deck be now?

 

That's a typical problem of calculating the equilibrium waterline of a boat (pontoon) loaded with a certain weight. It is not difficult but it is tedious if done by hand. There are many programs that perform those calculations, some of them free. In this forum there are many people who use these free programs who could advise you.
If you can export a 3D solid model in dwg format, in metric units, I could do these calculations for you.

 

Here it comes, see the attached files. For your convenience, the LCB (longitudinal center of buoyancy) values are measured from the transom of the pontoon.

When in static equilibrium, the longitudinal CoG of the pontoon+airplane has to coincide with the LCB. So, at the moment when the airplane is in the aftmost position (18 in = 0.46 m behind the transom), the overall CoG is situated at x=1.163 m. I have assumed 1200 lbs airplane weight (900 empty + the pilot + some fuel), and 1000 lb pontoon weight, equally divided between two pontoons. What you see in the attached calcs is relative to a single pontoon.

So, from the graph you can read that LCB=1.163 m is attained when the trim angle is approx. 5.5°, and the draft is approximately 0.37 m. The attached drawing shows how the pontoon will trim in that condition.

Cheers

 

Wow!! thanks so much for running those numbers for me. Really great to not have to build this just hoping that it will work out now!

 

I had actually calculated the COG of the raft and found basically the same number as you only I used the max gross on the plane, which made mine move back a little further.. I really appreciate you adding that table as now I can just use that and get my numbers. I am going to design the lift mechanism in a manner that ensures the tail will not get dunked in the water as the trim angle increases when the raft is lifting the plane. Now I will be able to do that perfectly! BTW I will make sure to show you guys pictures and videos when this is all done.

 

now that I have seen the calculations that daiquiri has run I would really like to be able to run some on my own for different positions and configurations if I need to. I think the table you have given me may be enough for really what I need but it would be cool to be able to run it myself. What Daiquiri did was exactly what I wanted to do. My only question is what would be my best option for a free program that will run those numbers?

I'm getting a solid works model of the hull sent to me from the airplane manufacturer that I am going to use to set on there. So I will be able to use daiquiri's numbers to put a plane in solidworks as the water surface and then ensure that I am not dunking the tail. The good news is that when the aiplane is resting on the water the wings basically have a 5° angle of attack and I want them to sit level when it is in the storage position. I am going to design the cantilever lift mechanism to tip the rear up that 5° right at the beginning of the lift motion so that it takes into account the trim of the pontoons raft initially. So initially the mechanism will just be sinking the raft until it gets to that 5.5° trim angle at which point the plane will be level on the cradle, with the water and the raft will be at its maximum trim angle. Then as the mechanism continues to lift it will be lifting the plane out with the hull basically in the same angle that it sits in the water, relative to the surface of the water. As the cantilever lifts it will be moving forward slowly (nature of a cantilever lift) which should begin to reduce the trim angle minimally, and then finally when the cantilever lift is at it's maximum position the assembly can then be rolled forward on the tracks which will center the load on to the pontoons giving the raft a 0° trim angle and the airplane's wings a 0° angle of attack.

 

I have used Freeship for this calculation: http://www.hydronship.net/index.php?lang=en . It is freeware and easy to use.

But I am pretty sure that even Solidworks could be used for this purpose, through following steps:
1) create a solid cylinder having the size of a single pontoon
2) create an inclined plane (at an angle A to the horizontal) which intersects the pontoon cylinder at some vertical distance from the bottom of the transom (distance T). This plane is the water surface. A is the trim angle and D is draft.
3) split the cylinder in two solid bodies, using this plane as a cutting tool.
4) enquire the mass properties (volume, CoG) of the newly created underwater body. These will be equal to the displacement and LCB, respectively, of the pontoon.
5) for each angle A, vary the draft D until you get a displacement equal to the load on the pontoon and read the final CoG position. This is the LCB of the submerged body.
7) Create a table of D and LCB vs. A.
That's it.

Regarding the wing angle in the storage position - IMO, perhaps it wouldn't be a bad idea to give it some negative AoA, in order to decrease wing-induced aerodynamic loads on the mounting and lashing points in case of strong wind gusts. Just thinking out loud...

Cheers