I admit it. I am really, really bad at physics. So please, don't rub it in, but do help me clarify a bit of confusion: Tramway Rd from I-25 heading east until it hits the Sandia Mountains in Albuquerque is about 5 km long and 330 meters up. I think about it a lot, because it sits between me and home, and I watch my FE plummet each and every time I drive it. So you will sympathize with my desire to drive that patch of road in the most efficient manner possible. It is often not congested, has no signals, and has a shoulder I can slide onto to let cars pass if need be, so I have a lot of latitude in picking my driving speed. Experience has taught me that slower is better. My MFD tells me the same, because if I try to drive much over 70 kph, the HV depletes and I end up with a high revving ICE that makes me shudder just thinking about it. And yet -- 54 kph is 15 meters/second, or 1 meter elevation gained per second Pulling the car up one meter requires 1350 kg * 1 meter = 1350 Joules of energy, So in one second, 1.35 kW of power. Doubling my speed then should double my power requirements for the elevation change, or 2.7 kW. The Prius ICE is reportedly efficient in the ~ 20 - 40 kW range. So why the problem with another 0.45 kW, to travel say at 20 meters/sec ?? (I did change the numbers just a bit to make calcs simple)
I think what you are missing is wind resistance. The faster you go the more of your fuel is spent just pushing through the air. At lower speeds it's less visible. So while the Prius EV might be good enough to lift your car those meters it also needs to push a lot of air out of it's way.
Managing your climb rate to deplete the battery to maybe three bars by the top of your ascent should actually improve mileage overall, because that gives you more capacity to store some of the energy coming back down. But no, you shouldn't feel like you have to totally punch it on the way up, even though air resistance at 70 kph is still a relatively minor factor. [kph, not mph...] . The ICE is actually at its best efficiency between 15 and 35+ kw output -- a rather flat curve, in fact. Have a gander at http://techno-fandom.org/~hobbit/cars/prius-curves.gif and think about adding a tach as a guide to how much you're requesting. I don't think crawling over mountains is a contributor to low mileage so much as just loafing along at medium speeds without letting the ICE shut down is. Thus, pulse-n-glide. . _H*
I think you're putting way too much thought into this. There's almost nothing you can do to improve your FE on a road like this other than to either go slower or avoid the route altogether. 5km is just over 2 miles...an almost insignificant distance even if you're FE is only 20mpg. I say pick a speed...prob. the speed limit. Set the Cruise control and let it do it's thing. Remember that your FE on the way down will be nearly equally as much better as it was bad on the way up....you can even use the gravity more to your advantage by trying to glide and allow speed to build as much as you're comfortable with.
A flaw in your logic is that the relationship between power and speed is exponential, not linear. Doubling the speed requires 2 to the tooth more power, not just double.
I admit that I've forgot most of my physics which I was pretty good at back in College (never gotten less than 95% on exams B) ). I think you're missing a few points here in your calculation. 1. You're mixing up potential energy with electrical energy. Electrical energy is first converted into mechanical energy, then stored as potential energy when climbing up hills. Of course, in the process, energy is lost due to friction and heat. 2. When coming down hill, the reverse happens. But still there's lost of energy. With that said, I'm no expert in what's the best way to drive through hills, but I tend to agree that using the battery on a up hill as much as possible to leave room for charge on the down hill may give you better results.
hyo: My writing was unclear -- that doubling of power requirement was only regarding the addition needed for the elevation change. Somerville: A 33% increase in speed from 15 to 20 m/s translates into 1.33^2 = 77% more air resistance -- around 2 kW at these speeds, I think. I'll double check when I find my PDA with W. Brown's simulator. Nowhere near enough to take the ICE out of it's 'zone'. Evan & Hobbit: I do go down after going up, but not immediately. A flat in between recharges the battery despite my wishes. I am not trying to manage the HV, but rather understand why the speed increase takes such a large hit on FE, which suggests to me that the ICE is inefficient at the higher speed.
No. As long as his speed is constant (beginning to end) there is no change to his kinetic energy. Yes. You are bad at physics. 1350 kg * 1 meter/second * 9.8 Newtons/kg = 13230 Joules/second = 13.23 kw
Hey, I know ABSOULTELY nothing about Physics, but I am from Albuquerque and know that stretch of road too well. Just went to my local (Santa Cruz, CA) dealership to get on the list for an 06.... :huh:
I am also in Albuquerque and live off of Tramway & Copper. I can't really help with any real world experience yet though because I am still on a waiting list. Scott
Scott and Judy -- this is great ! We'll knock those smug calis off their 'capital of the Prius world' yet Kiloran -- the car's 1350 kg is weight, and not mass, no ?
Actually, you should try to go up the hill on an empty battery. A lot of fuel is wasted hauling all those extra electrons up the hill.
Mass is defined in such a way that it is equal to the weight on Earth. When the interplanetary Prius arrives, let me know. For that, I'll pay over MSRP.
Nope. (Well, "sort of" if you are not an engineer.) Kilogram is a unit of mass. Newton is a unit of force or weight (i.e. gravity). The weight of 1kg under the gravitational attraction of the Earth is 9.8 N (Newtons). So the weight (downward force) of a Prius is actually 13230 Newtons (although lay people do not generally use Newtons). The power required to raise a mass of 1 kg by 1 meter is 9.8 Joules. To do so, repeatedly, at the rate of 1 m/s requires a power of 9.8 Watts. To do this for a Prius with a mass of 1350 kg requires 13.23kW.
It is reporting the vehicle's mass. The average man-on-the-street does not differentiate between the mass of an object and its weight and so, generally uses the terms interchangeably. Marketers use the terms that the average man-on-the-street will understand, in this case weight and kilogram. It's not correct but you understood what they meant.
I'll be damned. I have misunderstood that all these years So in the classic kid's question: If I weigh X Kg on Earth, how much would I weigh on the moon, the answer is: X Kg, because mass remains constant ? Jeez, I even remember the answer wrong. LOL. I finally get it. Weight's units are newtons, but common language reports mass. But when we go out to space, we continue to talk about weight, but switch over to newtons.
They are very different. It's just that the man-on-the-street doesn't know (or generally need to know) what the word "mass" really means. Mass is an abstraction, the resistance of a thing to having its momentum changed. Weight is a force and the force of gravity is a tangible thing. When the average man-on-the-street holds an object of 1kg in his hand, he feels its weight (~2.2lb or 9.8N) and calls its weight a kilogram. It's considered appropriate to equate the two in ordinary day-to-day situations but it's not correct to do so in physical equations.
Thank you Kiloran ! The man in the street can stay confused. As for me, I mass 60 Kg I weigh ~ 600 Newts And I dare anybody to prove me wrong