That last equation should have been W-h/N-mi:
Watt-Hour / Nautical-mile = 37.7 * (knots ^ 2)
From: electricboats@yahoogroups.com [mailto:electricboats@yahoogroups.com]
Sent: Wednesday, August 7, 2019 7:42 AM
To: electricboats@yahoogroups.com
Subject: RE: [Electric Boats] Re: New member
Mike-
This estimator is real close to your data:
Power = 37.7watt * (knots)^3
The cubic relationship between speed and power is significant.
Your data shows not 1kt/kw at low speeds, but instead:
0.3kw @ 2kts (6.6kts/kw)
1kw @ 3kts (3kts/kw)
2.5kw @ 4kts (1.6kts/kw)
Speed costs---you get to your destination proportionally, but the cost is power consumption as a cube of that speed.
The result:
Energy per distance = k * speed^2
In the case of your example:
KWH / n-mile = 37.7 * (knots ^ 2)
-Myles
From: electricboats@yahoogroups..com [mailto:electricboats@yahoogroups.com]
Sent: Tuesday, August 6, 2019 9:59 PM
To: electricboats@yahoogroups.com
Subject: [Electric Boats] Re: New member
The projected numbers are just about spot on from 20a and higher. Hard to do the slow speeds but if you use the rule of thumb of about 1kt per kW you see it is pretty close. We have done several Bayfields but not the 25.
The charger we normally use would be a 25a 48v (1200w) charger or a 15a 48v (720w) charger for smaller packs. The typical portable 2kW genset will drive 1600w
Mike Electric Yacht Pacific
Posted by: "Myles Twete" <matwete@comcast.net>
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