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Round Trip With A Prius And Trailer
Given a trailer that neither increases, nor decreases CDA, will the return trip down hill make up for the first trip uphill with a 500 lb trailer?
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In theory yes, but in practice, no. Reason being is the weight will require more energy to move uphill, and returns more energy going downhill. If the down grade is steep enough you’ll finding yourself using the friction brakes, turning all that “free” energy into heat and brake dust... best pot case is a grade gentle enough to allow you to DFCO down the other side and have the engine hold you back using no fuel... you’re still gonna have to brake at some point though...
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The cruise control on the Prius reverses the electric motors to slow the car and puts voltage back in the battery. I know I can't legally harvest that kinetic energy through gathering speed on the down hills, and I know there will be some loss of efficiency translating kinetic into electric. I was hoping for some experience from someone who had monitored their mileage both ways
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Yes the Prius uses regenerative braking, but at some point the battery will be full and you’ll have to use the friction brakes to slow down... if you had a long runoff at the bottom of the hill you could harvest the energy as kinetic energy (at risk of getting nailed for speeding) and then coast down over a long distance...
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If it's a standard Prius, the battery holds very little energy.
I've filled my 3.2 kWh Prius plug-in from empty to full coming down from Yosemite, and that was without a trailer. |
Unless it's a boat tail trailer, it will increase Cd and therefore CdA.
I've seen record trips with a small garden trailer, but that's towing with a 2ton van where the trailer is hidden entirely in the TV's wake. |
Mine will be hidden entirely in the Prius wake. It will be 4 feet wide and 40 inches tall.
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Quote:
M = mass of car m = mass of trailer T = angle of hill (assuming same angle up and down) d = distance of hill, and assuming duphill = ddownhill FD = aerodynamic drag force, assumed same for both FR,car < FR,car+trailer = force of rolling resistance You end up with the following work required for each situation to go up and down a hill of distance d at inclination T: Wcar = 2d(2Ma + 2FD + 2FR,car) Wcar+trailer = 2d(2Ma + 2ma +2FD + 2FR,car+trailer) The gravitational force does indeed cancel out, but it cancels out in both situations and you still have the force required to move the trailer around, whether uphill or downhill, which doesn't cancel. So, even if a = 0 it requires more work to get the car and trailer up and down the hill than the car alone, and thus more fuel. |
We are planning a round trip in July from Texas to Wisconsin, and back. I'll let you know the going, coming, and overall when we get back.
Previous trips with no trailer, but all our gear in the back of the car gave us 40 MPG. If I design the trailer skin correctly, I'm hoping to reduce drag on the rear of the car. |
Rolling resistance still increases either way. At max, aero drag is still only 80% of total drag on the car at 70 mph, and only goes down as speeds decrease.
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