T-Pain Is On A Boat: Re: [Electric Boats] Well, I am pleased to meet you, won't you guess my name.

Sunday, October 16, 2011

Re: [Electric Boats] Well, I am pleased to meet you, won't you guess my name.

Who said anything about needing 1000 lbs of motor? I just pointed out the numbers.

As you put it: Put down the calculator, walk over to the work bench. All the answers aren't in that little plastic box. I don't need to because I use to do marine wiring for a living. This includes starting with a bare hull and layout and setup all the systems. I have built a few of small charging systems for friends sailboats so they don't have to run their main engine to keep the batteries up. I have done a fair number of marine rewiring jobs and know whats the right way to do things. A lot of what I know comes from getting my hands dirty as well as my muzzle in a book and the classroom.

Now you talk about getting the most for the pint of gas... With your system it's not really that efficient. First you are taking mechanical energy and turning it into electrical energy then taking that same electrical energy and turning it back into machanical energy.

Your first step has a power loss of 35 to 45% going from your engine to your batteries. Your next step from your batteries to the water has a 20 to 30% power loss and that has nothing to do with the prop.

Now if you were to take the same input power from your engine and belt drive it to a prop and prop shaft your power loss would be 8 to 12%, with bearings and such another 4 to 6 %. Again this has nothing to do with the prop.

These numbers are real world numbers that are from testing in the real world by others. And can be found if you wish to go looking.

Am I getting useable voltage for for gas by the pint out of a package that weighs 80 lbs.? There are also better ways to get the same voltage using a lighter and more efficient package.

femm

--- On Sun, 10/16/11, Lochadio Who <lochadio@yahoo.com> wrote:

Am I getting every ounce of potential out of the alternators? not nearly.
Am I getting useable voltage for for gas by the pint out of a package that weighs 80 lbs.?

Absolutely.

I'm sure a 1000 lb motor would get more power out of the alternators, but the object of the exercise is pushing the boat.
You're talking about an extra 920lb of motor, plus fuel tanks(s) deck space and the extra hull to keep it all displaced.

With all due respect to your apparent education and expertise, you need to factor in the concept of diminishing returns.
Put down the calculator, walk over to the work bench. All the answers aren't in that little plastic box.

From: Femm <femmpaws@yahoo.com>
To: electricboats@yahoogroups.com
Sent: Sunday, October 16, 2011 8:51 PM
Subject: Re: [Electric Boats] Well, I am pleased to meet you, won't you guess my name.

Okay those would be the Gm-Delco Delcotron and they run from 43 amp to 68 amp output ratings at 14 volts, some were as high as 75 amps for single drive belt. Yes you maybe spinning four of them over. BUT you will not be able to pull full ratings out of all of them at the same time. If I was to put an amp-meter on each output lead it's very likely you are only putting out less then 25 amp per unit.

Lets do the math. 4 Hp X 500 watts= 2000 watts 2000 watts of power is all that you can get out of that gas motor. That also doesn't account for the power losses of your drive belt. I don't care if you are spinning one or ten alternators the total power you can pull is 2000 watts. That works out to 143 amps at 14 volts.

Yeah I can say I'm running a 10,000 watt genset with your 4 HP motor and in fact I can but you slap that system with a 4,000 watt load you A. slip the belt or B. kill the motor. Being able to spin over the alternators is one thing pulling their rated power out of then in something else. My big worm drive skillsaw will kill my friends 2000 watt genset if I try to cut to fast with it and the saw is rated at 1750 watts...