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Old 03-30-2011, 01:21 AM   #11 (permalink)
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Quote:
Originally Posted by some_other_dave View Post
That's where checking the BSFC charts comes in. From reading those, generally around 2000-2200 RPM and 70% load is the most efficient. The rest of the energy goes to noise and heat.

-soD
That part, I know. I know about BSFC and engine efficiency. It's the theoretical part I'm looking for. Forget about engines and cars for a minute. Think physics class. I know, it hurts.

Does it require more energy to create the same amount of kinetic energy? ball 1: 10mph in 10 seconds vs ball 2: 10mph in 5 seconds. Theoretically, I don't think it does. I'm just trying to confirm this so I can fully understand.

Getting back to cars: If all the difference in losses between accelerating faster and slower have to do with engine efficiency then there is no limit on how fast you accelerate for FE but the engine BSFC. I realize there may be other factors like increased friction.

I was under the assumption that the energy required to accelerate to X speed over time was exponential but I now think it's directly proportional. Just cant seem to find a source that can confirm this.

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Old 03-30-2011, 10:37 AM   #12 (permalink)
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Lets say you need 600 HP seconds of energy to get your car to 60 MPH, assuming all the other loss sources (aero, RR, mechanical friction in drive train, etc) were eliminated, then your source of that 600 HP-sec would be most efficient at its best BSFC (without time as a constraint).

600 HP-secs for 1 sec
300 HP-secs for 2 sec
100 HP-secs fro 6 secs

I can't see any reason for that to become proportional, however that is mostly instinct on my part, and could be refuted easily with a logical counter explanation.

It is basically changing the inertial state of an object moving in a linear fashion. I do stumble on the exponential cost of accelerating a flywheel, but have seen that just running an engine with no load at different RPM seems to use a linear volume of fuel.

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Old 03-30-2011, 05:43 PM   #13 (permalink)
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Today I did some quick tests.

0-40 then cruise engaged. Total time was 30 seconds. I accelerated at 3 different RPM.
With the CVT you can maintain a constant engine speed while accelerating.

3000 RPM=19MPG
2000 RPM=20MPG
1500 RPM=23MPG

Now the variable is total distance since the 3000 rpm acceleration was about 3 to 4 times faster than the 1500 rpm acceleration.

This means you travel less total distance when accelerating more rapidly, which increases the time you would cruise to the same distance. At 40 MPH this car averages about 45 MPG so I think it would be a wash overall if that was considered.

I guess I could do the test from point to point to equalize the distance, but then the 30 second reset time on the instrument (factory) in the car would not work since the time would be in between 30 second intervals.

Doing the test like this I just reset the meter and started off and the first reading was the figure I posted.

I would say within reason it probably makes little difference as long as you do not go to extremes like WOT. My car supposedly does 0-60 in 7.5-8 seconds with the 2.5 4 cyl and CVT.

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Old 03-30-2011, 09:20 PM   #14 (permalink)
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On this regard I think the only way to REALLY know the best way to accelerate would be to repeat the same trip in the same conditions day after day using the scan guage to show how much fuel over the current trip has been used in ML and how far you have traveled in meters (if it would do such a thing), procede to test a 1000m drive by accelerating hard to the same speed and go that distance, repeat with the different variables up to the same speed day after day until you determine what makes your vehicle use the least gas to go that distance.

Now any ideas how I get my SG to display how much fuel I've used so far in reasonably small units and likewise to display the current distance traveled on the current trip in reasonably small units?

I would like to log this with temperature to figure out what form of acceleration is really the best over time. BSFC doesn't always match reality.

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Old 03-31-2011, 07:18 PM   #15 (permalink)
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Quote:
Originally Posted by orange4boy View Post
That part, I know. I know about BSFC and engine efficiency. It's the theoretical part I'm looking for. Forget about engines and cars for a minute. Think physics class. I know, it hurts.

Does it require more energy to create the same amount of kinetic energy? ball 1: 10mph in 10 seconds vs ball 2: 10mph in 5 seconds. Theoretically, I don't think it does. I'm just trying to confirm this so I can fully understand.

Getting back to cars: If all the difference in losses between accelerating faster and slower have to do with engine efficiency then there is no limit on how fast you accelerate for FE but the engine BSFC. I realize there may be other factors like increased friction.

I was under the assumption that the energy required to accelerate to X speed over time was exponential but I now think it's directly proportional. Just cant seem to find a source that can confirm this.
KE is just related to speed, so in a vacuum/frictionless surface it doesn't matter how long it takes you to get to whatever speed, you'll use the same amount of energy to do it.

For cars in general the limiting factor is engine BSFC and for automatics tcc lockup since torque converters are really inefficient. The energy required to accelerate to some speed, assuming a manual trans or tcc locked most of the way is proportional to the speed squared. Since brakes are just devices that turn KE into heat, braking distance is also proportional to speed squared, eg someone's KE/braking distance at 70mph is about twice that of someone at 50mph.
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Old 03-31-2011, 09:02 PM   #16 (permalink)
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Tractive effort power looks like the road load force + the inertial forces times the velocity. The true aerodynamic resistance can't be ignored in the real world and can play a significant role in the over all resistive components of the Road load power =(rolling resistance*mass*gravity*cos(theta)[(rolling resistance term)]+1/2*rho*cd*0.8*af*v^2 (aerodynamic drag term)+m*g*sin(theta) [Grade resistance])*v.

Therefore the acceleration force looks like Pte-Prl as mentioned previously but the driveline efficiencies cannot be ignored or assumed at a single reference value, also the inertial term changes with each gear, the effective mass is not a constant, its highest in lower gears. I could go on, but that's enough for now
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Old 04-01-2011, 12:03 AM   #17 (permalink)
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Pretty much. Rolling resistance and the drag coefficient should be expressed as polynomials through interpolation from actual data in addition to inertial forces and an expression for the efficiency of all the driveline components, but all of that data is a pain in the ass to come by. I imagine inertial force should even be a function of whether or not the tcc is locked.

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