3 ms·

TO answer everyone's questions below, think about the following example: You have one car that's 1,000 lbs and one that's 10,000 lbs. When you start each car,

TO answer everyone's questions below, think about the following example: You have one car that's 1,000 lbs and one that's 10,000 lbs. When you start each car, it will take a lot more energy to bring the 10k lb car up to the same speed as the 1k lb car, but after that, the effects of drag will weigh more heavily on the lighter car (F=ma, so the rate of deacceleration is a lot higher for a smaller mass), causing constant fluctuations in the motor shutting on and off. With the the 10k car, the transitions would take much longer. The time in between accelerations would be longer, and the motor would be on for longer spurts. You gain a great deal of efficiency for simply leaving a mechanical system running vs constantly modulating a mechanical system (i.e. shutting it on and off). This would work in favor of the heavy car being more efficient. Of course, the heavier your car is the worse your 0-60 split would be, so for a performance car that's a negative. Also, the heavier car would experience greater bearing resistance + friction. This is really difficult to quantify, and my gut is that it's outweighed by the momentum advantage. But overall, for a car that ways 2k lbs, a 5% increase in mass is pretty negligible, especially for casual highway and local driving.

function_seven 13y ago
> but after that, the effects of drag will weigh more heavily on the lighter car (F=ma, so the rate of deacceleration is a lot higher for a smaller mass), causing constant fluctuations in the motor shutting on and off. All other things being equal, a heaver car will experience the exact same aerodynamic drag as a lighter car. Only the cross-sectional area and shape of the body come into play when determining how much opposing force air resistance provides. By the same token, the motor must supply a constant amount of torque to the wheels to counteract that force. If the force remains unchanged, so too does the torque required to counteract it. You're right about the rate being different, but as far as I know the motor doesn't cycle on and off like an air conditioner compressor does. It maintains a steady output. What we're left with is an additional component of drag due to rolling resistance. A heaver car will deform the rubber tires more than a lighter one, and put more normal force into the bearings, causing that rolling resistance to be higher. I'm not sure what fraction of the total drag is due to rolling resistance vs. wind resistance--like you, I bet it's comparatively small, but still significant enough to show up in reduced range. > Also, the heavier car would experience greater bearing resistance + friction. This is really difficult to quantify, and my gut is that it's outweighed by the momentum advantage. I may be wrong here, but I don't see having an increased momentum as being an advantage. It takes more energy to build up that increased momentum, and you don't get it all back from regenerative braking A heavier car may be more efficient, if you calculate efficiency as [Energy used] / [Total mass], but the better way to calculate the real-word efficiency is [Energy used] / [Mass of stupid driver who doesn't deserve this car as much as I do, when do I get to have one?]
- squigs25 13y ago
  Sorry - was away for a while. The drag FORCE will be the same, but the lighter car will deaccel faster as I clearly stated above. F=m*a thus: F/m = a Think of it like this: it's really easy to push a car that weighs one pound. Much harder to push a car that weighs 1000 pounds. The momentum advantage was explained in the first point.

function_seven 13y ago

> but after that, the effects of drag will weigh more heavily on the lighter car (F=ma, so the rate of deacceleration is a lot higher for a smaller mass), causing constant fluctuations in the motor shutting on and off. All other things being equal, a heaver car will experience the exact same aerodynamic drag as a lighter car. Only the cross-sectional area and shape of the body come into play when determining how much opposing force air resistance provides. By the same token, the motor must supply a constant amount of torque to the wheels to counteract that force. If the force remains unchanged, so too does the torque required to counteract it. You're right about the rate being different, but as far as I know the motor doesn't cycle on and off like an air conditioner compressor does. It maintains a steady output. What we're left with is an additional component of drag due to rolling resistance. A heaver car will deform the rubber tires more than a lighter one, and put more normal force into the bearings, causing that rolling resistance to be higher. I'm not sure what fraction of the total drag is due to rolling resistance vs. wind resistance--like you, I bet it's comparatively small, but still significant enough to show up in reduced range. > Also, the heavier car would experience greater bearing resistance + friction. This is really difficult to quantify, and my gut is that it's outweighed by the momentum advantage. I may be wrong here, but I don't see having an increased momentum as being an advantage. It takes more energy to build up that increased momentum, and you don't get it all back from regenerative braking A heavier car may be more efficient, if you calculate efficiency as [Energy used] / [Total mass], but the better way to calculate the real-word efficiency is [Energy used] / [Mass of stupid driver who doesn't deserve this car as much as I do, when do I get to have one?]

squigs25 13y ago
Sorry - was away for a while. The drag FORCE will be the same, but the lighter car will deaccel faster as I clearly stated above. F=m*a thus: F/m = a Think of it like this: it's really easy to push a car that weighs one pound. Much harder to push a car that weighs 1000 pounds. The momentum advantage was explained in the first point.