I was just sitting around trying to figure out the significance of the little car icons on the consumption screen and came up with this. Assume car mass = 1,311 kilos (2890 lbs) velocity = 96.6 km/hr (60 mph) = 26.8 m/sec From Kinetic Energy = (1/2)*m*v^2 K.E. = (1/2)*1311*28.8^2 = 5.44x10^5 Joules One "car" is 50 Watt-Hours or 50x3600 = 1.8x10^5 Joules So at 60 miles/hour the car's K.E. is about 5.44/1.8 or 3 "Cars". From early indications (only got 100 miles on the car) I'm getting around one "Car" for each time I decelerate from 60 to a dead stop, so the regenerative efficiency is around 30%. Anyone know if this is in the ball-park? Just musing....
I lack the skills necessary to offer an independent assessment, but I did find a Toyota Hawaii web site that comes to much the same conclusion as you: http://www.toyota-hawaii.com/vehicles/Priu...us/braking.html
Yep, it is. The KE equation also illustrates the huge energy penalty involved with even minor increases in velocity. No matter what vehicle you drive, and we'll also ignore the huge aerodynamic drag penalty for a moment, a vehicle operated at 80 MPH has to use much more energy - and get much lower fuel economy - than a vehicle operated at 55 MPH. This is also why it's much easier to manage crash forces at 55 MPH than at 80 MPH, and why high speed crashes usually have a bad outcome.
If you ignore drag, it is getting to 80MPH that uses energy. Maintaining 80MPH uses none on a level surface. An object in motion stays in motion, an object at rest stays at rest until changed by an outside force. However, mechanical friction as well as air resistance would be that force working to put the object at rest, which must be counteracted.
Also - again ignoring aerodynamic drag and friction with the road surface - relatively minor increases in speed require proportionately larger energy expenditure at higher speeds. To accelerate from 30-55 MPH takes less energy than to accelerate from 55-80 MPH.
I reckon one green leaf-car of energy recovered is equivalent to the energy provided by about 1 tablespoon of gas. See Prius: gas equivalent of green leaf-cars Of course it is always better to touch the brakes as little as possible; it's much better to not lose your kinetic energy in the first place than it is to recover some fraction of it. What the Prius does is to get some but not all of it back.
This also helps to illustrate why "impulse" driving is most efficient for the Prius (i.e. quick accleration to speed and then cruise at constant speed as long as possible). Someone did measure the regnerative amperage and found that light braking at about a 2 mph/second deceleration is most efficient for recovery of energy(plus it doesn't make people behind you as mad).
"This also helps to illustrate why "impulse" driving is most efficient for the Prius (i.e. quick accleration to speed and then cruise at constant speed as long as possible)." Does anyone have actual data to prove this?
i read somewhere that energy recapture was more on the lines of 8-15%. also i find it hard to believe you can determine this because the rate deceleration would play a part in this. to maximize regen you would have to coast to a stop and then terrain would be playing a major role. a lot of things to consider and i think you will find that maintaining a car per deceleration wont be likely. i have seen 4 cars before but that was from stop and go traffic where i never really got over say 25 mph.
I've never seen anything above 3 cars. But on a test drive of the RX400h, the last 5 mins had 4 "E" squares. Weird
As a point of interest, Honda claim to be able to recapture >95% of kinetic energy via regenerative braking in the Accord hybrid. http://www.greencarcongress.com/2004/09/un..._the_hood_.html
95%? hard to believe... actually if you read it, it says 95.2% of the kinetic energy is available to be converted to electricity... which is good...that only means that 4.8% is lost through drag and friction losses from the tires, transmission, etc... at least that is what i take it to mean... Honda claiming anything else is simply misleading advertising
DaveinOlyWA, note how Honda says is "available" for conversion. Since most genrators-motors run at about 25-35% efficiency you are only going to "recover" 25-35% of the 95.2% "available". Hence the basic shortcoming with electricity generation from a mechanical (motor-generator) system is that at best about 35% can be rcovered.
<div class='quotetop'>QUOTE(rwlade\";p=\"89635)</div> Mass should be in kg, right? I though "lbs" reflects weight of a body with gravity acting upon it. Also, I believe "velocity" is actually "instatneous velocity", a scalar value. Regular old "velocity" is a vector, with both direction & magnitude.
<div class='quotetop'>QUOTE(200Volts\";p=\"90554)</div> Most generators / motors are typically more than 95% efficient, and some are more than 98% efficient. If they were only 25% efficient, 75% of the electrical energy going into them would get converted into heat, and they would melt! The bottle-neck in recapturing regenerative energy is not the motor/generator or even the power electronics (which are also very efficient) but the battery itself. The bigger the battery, or the more current it can accept, the better. The Prius battery can only handle a relatively small current input (21kW max) compared to what braking events can provide, so most braking cannot be achieved with regen alone - hence a lot of what is available is lost because the battery isn't able to absorb it all. GMs EV1 had a big battery capable of soaking up a lot higher regen currents, and they were able to show a 44% increase in urban range by switching on regen, which I think is incredible as I wouldn't have thought that much time is spent accelerating or braking even on the urban cycle.
<div class='quotetop'>QUOTE(jayman\";p=\"89765)</div> Ack! Now you've gone & done it! It's really not good to ignore air resistance -- mother nature gets quite upset & steals a lot of your power if you do! Look at this: Power required to maintain a given speed = 8.702 x 10^-6 x Cd x A x V**3 where Cd = coefficient of drag A = square feet of frontal area V = Velocity, in miles per hour For the prius, Cd = .26, A = 24 sq. ft. so, power required to maintain various speeds: 35 mph -- 2.3 hp (1.7 Kw) 55 mph -- 9 hp (6.75 Kw) 85 mph -- 33 hp (24.75 Kw) 134 mph -- 130 hp (97.5 Kw) -- Wasn't 134 mph the speed record they set in a prius? ( 1 hp = 750W = .75 Kw) Just to beat wind drag & rolling resistance. The 8.702 is a fudge factor that can be argued, and will change with different tires, inflation pressures, altitudes, etc, but the point carries -- power required goes as through the roof as velocity increases. It won't change by a whole lot in any event. FWIW, dave.
<div class='quotetop'>QUOTE(chrism07924\";p=\"90714)</div> Mass should be in kg, right? I though "lbs" reflects weight of a body with gravity acting upon it. Also, I believe "velocity" is actually "instatneous velocity", a scalar value. Regular old "velocity" is a vector, with both direction & magnitude. [/b][/quote] Yes, the calculations were done in the MKS system. Velocity (vector) and speed (scalar) can (and usually are) functions of time. Your use of the word instantaneous implies time dependence and that is true for both the scalar and vector variables. Don't take my musing to be a complete dynamical analyses of the system, I only was trying to get a feeling for the significance of the little car icons.
<div class='quotetop'>QUOTE(dstrout\";p=\"90789)</div> Great post, Dave. Intuitively, the velocity cubed relationship makes sense. Where did that equation come from?