Catamaran Hull Type Power Calculations and more

Hello everyone, so I am working on a catamaran boat which serves the purpose of cleaning a lake from it's plastic waste etc... basically a lake cleaning robot to be more clear. And after designing the model i now want to find the power needed to power up the system, I did research about it but all i was able to find were not as beneficial to me since it's not in the metric system. i want to find the power equation to calculate it for this hull with dimensions of : 100cm length, 40cm height, 70cm width. And i would like to also ask how am i supposed to do quantitative assumptions for such a thing?
some other things that i would love to get some insight about which are :
how can i determine the thickness of the material am going to use ? " material that i am thinking of using is PETG with an average density of 1.27 g/cc.
how can i determine thrust of the motor i will choose " after determining the power"
(below is attached the design and also some calculations that i once tried to do but then found that i cant be using inches and then continue with metric system, i should be sticking to one system for the accuracy of these calculations)

 

Welcome to the Forum Abdul.
Is this a university project / exercise or do you actually intend to build a boat to clean waste from the lake?
Have you started to build the model yet?
Do you have any other drawings showing how the boat will clean up waster - I presume it will have some sort of scoop on the front?
I wonder why your transoms are simply 'chopped off' - relatively the hulls will have a bit more resistance like this, when compared to if there was rocker in the hull bottom.
I am intrigued by your calculations - do you really need 2 kw to drive a 1 metre long lightweight catamaran?
And do your electric motors really work at 28,000 rpm?
What speed of rotation do you anticipate the propellers will be turning at?

 

Welcome Abdul,

What voltage are you running at 28,000 RPM's?

 

Quick and very approximate numbers for displacement craft.
Displacement speed (knots) = 1.2 * SQRT( Length Waterline (Feet))
At that speed normal propulsion is about one horsepower per ton of displacement.
The power number here can be +/- 50%, but clearly far less power than your calculation.
The propeller should be turning maybe 1000 RPM, maybe 2000, but again, far less than your number.
Is this craft towing or pushing something relevant to the calculations? Please give details.

 

well as a first, isnt there an equation but with the metric system ? " meters "
second, i honestly followed this one video on youtube if you would like to take a look at

and i used the same path of calculations as the person in the video did.
the main objective of the design is to collect waste from the surface of a lake by using a net attached to the back of the boat which will be attached by using the hooks on the back

 

thank you for reaching out kind sir, actually this is a university project that if all things go well can be anticipated into my actual graduation project by next year.
attached are more drawings of the design and as for the cleaning mechanism it'll be in the form of a net attached to the back of the hull by hooks where the boat will walk through the waste in order to collect it in it's net.
the "chipped off" is a design reinforcement way to balance the model ( all done in solidworks ) but it isnt much of a necessary thing. as for my calculations i followed a video from youtube that explains how to choose a brushless motor for an rc boat () and did all the calculations accordingly. the boat has a unloaded mass of 15 kg and we are putting a limit of waste mass on this boat to be around 15 kg also so a total weight of this boat loaded with waste is around the 30kg line, i would love if you can provide me with better calculations in order for me to work better on this "preferably metric system values" and i again thank you a lot for taking your time

 

am running around 22.2v from my batteries

 

Abdul,

Thank you for your response.
I watched the video and have some R/C experience.
The video is misleading for you because you have a much longer "hull-length" than just the boat.
Your gear you're towing is considered part of your boat length.
In addition, your boat weight varies by how much debris you've collected.
So, you have a hugely varying boat weight depending how much debris is in the catchment system.
Also, your boat speed is going to be much, much lower than the R/C enthusiast's.
I suspect you're going to need a much, much lower prop RPM and of a huge diameter.

This is not an R/C forum.
You may get lucky and find someone on here with lots of R/C experience, it's not me.

Can you show your towed gear and how it attaches?
Also how it loads and empties, max and min weights of debris.

What is your design speed?

I don't want to discourage you but an R/C forum may be better suited to help you.

BB

 

No no absolutely I understand your approach; I am not mainly looking for an r/c related stuff but more of a catamaran hull type calculations.
the net that will be attached to the back isn't fully designed yet unfortunately, but here are some information that might be helpful to gain the better idea
the designed speed is around 1-2m/s and the catchment system is basically a wired net which will be attached to the hooks at the back and it will be loaded when the boat drives through the waste ( since it's a catamaran type which has an empty space in the middle) and this will directly path these waste inside the net and the net will be loaded accordingly. I can tell the components specifications are a bit overqualified but I thought to myself that it's okay since the boat will be autonomously running and I would need less maintenance since the batteries are in bigger capacity and the motors can run for the desired speed hence i wont put too much load onto them

 

Hi Abdul,
regarding your calculation in your opening post, do you understand, that a current draw of 110 A from a battery of 3.3 Ah capacity results in an empty battery after 0.03 h or 1.8 minutes? (Just a theoretical raw number, without regarding a lot of effects.)

 

I actually didn't know that; do you mind elaborating the calculation used to come up with this and also, I wont be using the motor to its full power so that's something I can share with you as well. 1 or 2m/s is what i am going for and do you mind suggesting a better battery option ?

 

That's very basic electric around Ohm's law.
If you take a Li battery you can mostly draw 70 per cent of its nominal capacity without shorting its life. So if your battery is rated 3.3 Ah (Ampere hours), 70 % of it is 2,3 Ah. The latter is the really useable capacity assumed you draw a current of only 2,3 A for one hour. Sometimes the nominal capacity is rated related to 0.1 C, which means 230 mA only for ten hours. Read about Peukert's law.
https://en.wikipedia.org/wiki/Peukert's_law

 

well let me ask this then if you don't mind sir, i've upgraded the battery choice to a 6200mAh 6s with a C rating of 40, and i calculated the full load current of my motor which was 13.19 Amps and from there i would like to ask if usage of this battery is going to be efficient since i am only going to use around 1.5 percent of the motor full speed
full speed was found by using : V = 2*pi*r ( r is the radius of the propeller which is 0.045 m )* N / 60 ( N is the rpm which accordingly i was able to chose the motor 28000 rpm)
and the desired speed is only 1.5% of that hence the usage of the motor is only going to be 1.5% of it's full power. and from that the power consumption is also lowered simultaneously

 

6200 mAh = 6.2 Ah ; 6.2 Ah * 70/100 = 4.3 Ah usable capacity
current 110 A means: 4.3 Ah / 110 A = 0.04 h = 2.3 minutes. So battery is empty after around 2 minutes.

The nominal rpm of such RC motors is measured with no load at all, free spinning rotor. If you fit a propeller on it and operate it in the water rpm will be about 30 % of the nominal value (this depends heavily on the conditions).

The result of your formula V = 2*pi*r * N / 60 is not a velocity but the lenghts of the path which one of the propeller blade tips is decribing in one second (multiple circumferences of the circle).

You can not simply suppose that your 28000 rpm motor, reduced to 1,5 % of its rpm will be able to turn the boats prop in real conditions. It needs a minimum of torque for this.

Plastic debris in a net towed by a boat may not weigh much but produces a huge drag because of the "irregular forms" and the random orientations of the debris pieces (e.g. cups with opening ahead and such things). This must be included in the calculations regarding the boats drive.

 

I propose you here attached an alternative design within your dimension (L 1,0 m x B 0,7 m x H 0,4 m) and targeted weight (30 kg), with a good dose of rocker so that to reduce the immersed transom area and its related drag. I give the hydrostatics for 4 drafts covering displacement from 15,7 to 38 kg. And also a preliminary estimation of the drag and of the power estimation (for D 30 kg) using "SA-VPP power catamaran" application, althought the conditions of validity are not fulfilled (for Froude > 0,4 part of the application). More on the application here :
SA-VPP power catamaran | Boat Design Net
Anyway, for the drag better knowledge, and before fixing the power to install, I strongly recommend you to test at scale one your catamaran in operative conditions, which should not be very difficult in any standard towing tank or calm water area due to the small size and weight. To note that the preliminary estimation shows a high jump of power from 2 to 3 Knots (Froude 0,33 to 0,49).