45psi matter? CANbus data turning me into a drag-queen..er..

[sparky]

... king... duke?

This started as an exercise to determine if the above-spec tire pressures I've been running at (45psi) are really saving any kWhs or just making the so-so grip/handling of these Ecopia low rolling resistance tires even worse.
For those who don't want to read the rest; bottom line: I'm going back to setting my tire pressure to 35 psi. For the rest, sorry this is so long and please, if any of you hypermilers out there see problems wth my methods here feel free to correct. I'd love to learn more.

Awhile ago, I posted some data from my CANbus reader that I interpret as RPM and described my process for scaling that RPM data to vehicle speed. Since RPMs give us the LEAF speed and it gets reported about 50-100 times /sec we should have a precise way to measure the acceleration or deceleration of the LEAF. The LEAF is direct drive, RPM is still reporting motor (and wheel) turns when in neutral (unlike my ICE).
So, when we pop the car into neutral and record the deceleration, what we see is the result of two main forces acting on the car (per Wikipedia).

F = { Crr * M * g } + {Cd * A * 1/2 * rho * V^2}
F = { rolling drag } + { aerodynamic drag }

I don't like these forces since they sap my battery and give some of us what the French call inquiétude de distance or what GM has trade-marked as "range anxiety".
But that's physics.

When a car accelerates, these forces look like the plot below; with the force due to aero drag increasing as velocity-squared and force due to rolling resistance staying fairly constant.

http://www.atmosphere.mpg.de/enid/Information_ss/Velocity___air_drag_507.html

F = Ma so with some algebra: a = { Crr * g } + { Cd * A * 1/2 * rho V^2 } / M

a = dv/dt or just slope of the velocity over short intervals.
So, with my CANbus reader, I can get from LEAF RPM, to dv/dt and therefore a.

We have equations and we need some of the constants.
MKS units here (meters (m) kilograms (k) seconds (s)
M = mass = 1605 = mass of LEAF 1525kg + mass of driver 80kg
g = gravity = 9.81 m/s^2
Cd = coefficient of drag = 0.29
A = frontal area of LEAF = 2.27 m^2
rho = the air density = 1.22kg/m^3 (I adjusted for alt & temp to 1.15 using CRC tables)

RPM to m/s: GR = gear ratio of LEAF (1.79377) tire circumference: 2.05m
So RPM to m/s = Rev/minute * 1/GR * 1 minute / 60sec * TireCircumference m/rev
RPM_2_V coeff = 0.004306 (min * m )/(sec * rev)

So, we use the RPM/sec data to determine a for a given RPM (scaled to velocity, V).
We want to solve for Crr which is supposed to show a change vs tire pressure.
Although we know Cd, I decided to check my units and algebra (and physics) by solving for Cd using the 2 equations & 2 unkowns (Crr & Cd) at two widely differing velocities: ~75 MPH & ~ 15 MPH (from m/sec)

The stategy here is to take a data set at high speed and then a data set at low speed with tire pressure going from 45 psi for one high/low test set and the 35 psi for the other.

Using the two speed tests we can solve for the Crr (and Cd). Ultimately, we will just use the dv/dt number of the low speed 45/35psi tests to determine if the Crr changes sufficiently to warrant 45 psi vs 35 psi.
Sooo, this is really a simple test that I've complicated for the sake of completeness.

As an example of raw data, here's a plot of RPM taken on one of the high-speed test runs.
You can see my LEAF accelerate up and then suddenly decelerate when I let off on the pedal.
Then it decelerates strongly due to regen until I pop it into neutral. Then it flattens out nicely and we can compute a slope (dv/dt) with a 2-4 secs of data. I run this test going both ways on the same road to cancel out the slight slope in road. No wind, about 75 deg, dry surface, typical SoCal mid-morning.



So, here's my attempt at solving for Crr and Cd using raw dv/dt data from the LEAF CANbus.

Lest we forget: a = { Crr * g } + { Cd * A * 1/2 * rho V^2 } / M
High speed dv/dt: 0.37904 = Crr * 9.81 + Cd * 0.8916 (using V = 32.7m/s)
Low speed dv/dt: 0.1328 = Crr * 9.81 + Cd * 0.044613 (using V = 7.32 m/s)

From these we get Crr = 0.01221 and Cd = 0.2907
The Cd number is quite heartening to see ( I actually exclaimed "holy s**t this CANbus s**t works!) and tells me my method's not too bad. I still think it's a bit low since I wouldn't expect to come that close to the calculated number.
Probably due to my estimation of rho (air density).

Ok, almost done. Crr is 0.01221. From other LRR tire data I expect these Ecopia's are about 0.007?
Anyone got a published value? The rest is most likely due to the brake pad friction, transaxle fluids, etc.
The key is, what does this number do for energy loss and how does it change when we change tire pressure (even I almost forgot the point of this whole exercise).
Well, that's fairly easy.
If we go back to the Force equation at the top:

F = { Crr * M * g } + {Cd * A * 1/2 * rho * V^2}
F = { rolling drag } + { aerodynamic drag }

Force times distance = work = energy.
So, if the Crr goes up, the force against the car goes up and the energy loss goes up.

F * d = Joules * 1/3600s/hr * 1/1000 W/kW = Energy used in kWh

For example from the measured a value (0.37904 m/s^2) at high speed.
Total drag force F = Ma = 1605kg * 0.37904 m/s^2

If we travel a distance of 100 km = 100,000 m then we'll use
1605 * 0.37904 * 100,000 m / 3,600,000 = 16.9 kWh used to go about 62 mi @ 75 MPH.
That's just the loss due to drag, inverter and other non-drag losses give us maybe 85% efficiency?
Then throw in hills, headwinds, etc.
So, 16.9/0.85 = 19.9 kWh of battery use. I think that's plausible.

So, plug in the two values of Crr I calculated using the same drive cycle at 45 psi vs 35 psi and I get a whopping change to Crr from 0.0122 to 0.0125 which yields a change of energy use over 100 km of about 170 Wh (also it's within my error bars so hardly convincing).

So, to recap; I used the CANbus 0x1DA fields to get highrate RPM vs time.
Computed dv/dt at slow speed to estimate drag dominated by Crr.
Tested at 45 psi and 35 psi and dv/dt barely registered a difference.
Ran through the numbers on this small difference and believe the kWh diff between battery LEAF charging cycles makes it too small to bother with the higher psi levels I've been keeping my tire pressures at.

My theory for this is the LRR tires are already pretty elastic at 35 psi and increasing to 45 psi just doesn't yield the same benefit as non LRR tires.

BTW, has someone posted 0-60 times using the CANbus data?
If not, I'll try and put something up soon.

 

[SanDust]

Interesting information. You really can't derive some of the variables (like Cd) using your methodology because you're missing the drive train losses which can be significant. (For sure more significant than rolling resistance at higher speeds). But of course you can just do a comparison with different tire pressures. Without knowing the actual numbers, you'd expect tire pressure wouldn't matter much given that Nissan engineering had every incentive to maximize range, could have specified a higher psi, and didn't.

Just curious if inflating to 35 psi rather than the recommended 36 psi for testing was intentional or, as I'd assume, just a slip of the keyboard?

 

[Herm]

45psi is quite mild, hypermilers start talking at 70psi.. and true it will never be as big a contributor as speed.. thats why you never see hypermilers exceeding the posted speed limit, and quite often staying below it. Every little bit helps.

Now you need to become an ecomodder, but those are really weird people

http://ecomodder.com/

 

[sparky]

Now you need to become an ecomodder, but those are really weird people

http://ecomodder.com/

Thanks, but I think I'll pass on becoming an ecomodder, my wife thinks I'm weird enough.

Interesting information. You really can't derive some of the variables (like Cd) using your methodology because you're missing the drive train losses which can be significant. (For sure more significant than rolling resistance at higher speeds). But of course you can just do a comparison with different tire pressures. Without knowing the actual numbers, you'd expect tire pressure wouldn't matter much given that Nissan engineering had every incentive to maximize range, could have specified a higher psi, and didn't.

Just curious if inflating to 35 psi rather than the recommended 36 psi for testing was intentional or, as I'd assume, just a slip of the keyboard?

Actually, I failed to RTFM on the spec psi, thanks for correcting.
Can you explain what you mean by missing drive-train losses? Are they another force against the forward motion?

 

[ttweed]

RPM to m/s: GR = gear ratio of LEAF (1.79377) tire circumference: 2.05m...

Wow! You definitely have too much time on your hands, sparky!

My only question on your methodology would be whether inflating the tire to higher PSI actually changes the effective rolling circumference enough to effect your calculation? I see only one assumed value of 2.05m being used. Did I miss something? Perhaps the elasticity of the tire results mostly in an increase in section width rather than circumference when PSI is raised, but if there is any actual increase in rolling diameter at higher inflation levels, wouldn't that have to be accounted for in your comparison to be entirely accurate (apples to apples)?

TT

 

[SanDust]

Can you explain what you mean by missing drive-train losses? Are they another force against the forward motion?

The chemical energy doesn't all make it to the wheels. You have losses in the motor, gearbox, bearings, and so forth. Those will be larger than rolling resistance losses at low and high speeds, and probably in between.

 

[kevin672]

Can you explain what you mean by missing drive-train losses? Are they another force against the forward motion?

The chemical energy doesn't all make it to the wheels. You have losses in the motor, gearbox, bearings, and so forth. Those will be larger than rolling resistance losses at low and high speeds, and probably in between.

I think the point is that the calculation for the forces due to friction and drag are fine, but any calculations for the energy required to overcome that force will be higher because of losses due to the inverter efficiency, electric motor efficiency, and reduction drive efficiency (there's energy loss at each of those steps). Putting the car in neutral removes the inverter and motor, but not the friction in the reduction box, wheel bearings, etc.

 

[DaveEV]

Most of the ecomodder guys highly suggest doing A-B-A tests to cancel out inter-run variations as quickly as possible. The more you repeat the tests, the more you can be sure of the data, of course.

For example, one possible influence on your A-B test here is that run A has warmed up your gear fluids and axle grease - thus giving your B run an advantage. Also, when testing tire pressures - one has to remember that the tire will heat up as you drive - so at a minimum recording before/after pressures should be informative.

Also, typically a coast-down test in one direction is usually sufficient - but since if you have to go out and back anyway, more data is usually better. I would suggest not averaging the run data, but instead comparing them independently.

 

[evnow]

So, to recap; I used the CANbus 0x1DA fields to get highrate RPM vs time.
Computed dv/dt at slow speed to estimate drag dominated by Crr.
Tested at 45 psi and 35 psi and dv/dt barely registered a difference.
Ran through the numbers on this small difference and believe the kWh diff between battery LEAF charging cycles makes it too small to bother with the higher psi levels I've been keeping my tire pressures at.

My theory for this is the LRR tires are already pretty elastic at 35 psi and increasing to 45 psi just doesn't yield the same benefit as non LRR tires.

Very interesting - exactly as I'd expected. Unfortunately many of these remedies for hypermiling have become old-wive's tales, rather than fact based.

 

[Electric4Me]

To be fair, similar comparison tests should be performed for other aspects of the car's handling on the tire pressures of 35 (36) and 45. On my Subaru, I prefer the handling of the tires at 45. (this is certainly a tire & car specific variable)

I'd be really curious on the results of the same coast down test you performed on non-LRR tires. Have any one to trade with for a test? I've been wanting to test fit my friend's WRX STi wheels and do some testing but haven't had the chance.

Thanks for posting Sparky!

Bill

 

[DaveEV]

Very interesting - exactly as I'd expected. Unfortunately many of these remedies for hypermiling have become old-wive's tales, rather than fact based.

FWIW - looking at the data over at EcoModder, the data indicates you'd probably see around 1.5% efficiency in going from 30-40 psi (probably on a car with an OEM setting of 32 psi) - so I'd expect less than that going from 36 psi - 44 psi. Let's say it's 1% - or less than a mile in most situations using back-of-the-envelope calculations.

Compared to sparky's calculated results of 170 Wh over 100 kM - assuming that you're getting 4.0 mi / kWh on the dash (and an estimated range of 84 miles before turtle), that correlates to about 230 Wh (84 mi / 100 km * 170 Wh) or 0.9 miles of range gained over a full charge. Which happens to be within a tenth of a mile using my back-of-the-envelope calculation.

Now - do you really want/need that extra 0.9 miles per charge? 99% of the time - probably not. But given the number of times people have either driven into the garage on turtle or pushed their car into the garage - it's probably not a bad idea.

Now we just need someone to measure 60-0 braking distances at 36 psi and 44 psi and get on a skidpad to measure g-fores so we can see if there's a difference there as well.

As for myself - I will continue to run tire pressures of slightly above 40 psi - the car simply feels more secure at those pressures to me. This also gives allows me to worry less about checking tire pressures since they will take longer to drop below 36 psi.

 

[TomT]

I have a Valentine G-Analyst and could do that for braking... A skid pad would be a little more problematic...

Now we just need someone to measure 60-0 braking distances at 36 psi and 44 psi and get on a skidpad to measure g-fores so we can see if there's a difference there as well.

 

[sparky]

Can you explain what you mean by missing drive-train losses? Are they another force against the forward motion?

The chemical energy doesn't all make it to the wheels. You have losses in the motor, gearbox, bearings, and so forth. Those will be larger than rolling resistance losses at low and high speeds, and probably in between.

True, but I'm trying to follow this part of your response:

You really can't derive some of the variables (like Cd) using your methodology because you're missing the drive train losses which can be significant.

It seems in order for me to solve for Cd (reasonably well) I just need "F" and "V" at two independent values.
F = { Crr * M * g } + {Cd * A * 1/2 * rho * V^2}
F = { rolling drag } + { aerodynamic drag }

So, Crr contains all the non-aero "pushing against the car" forces, no? My solved for value is 0.0122 about 50% of which I assume is the LRR tires.
The other electrical losses don't matter for this calculation since they're not a force against my motion. No?
I'm coasting, the car could just as well be "off" for this test (except I'd get no RPM data).
I don't see why I can't solve for Cd.

My only question on your methodology would be whether inflating the tire to higher PSI actually changes the effective rolling circumference enough to effect your calculation? I see only one assumed value of 2.05m being used. Did I miss something? Perhaps the elasticity of the tire results mostly in an increase in section width rather than circumference when PSI is raised, but if there is any actual increase in rolling diameter at higher inflation levels, wouldn't that have to be accounted for in your comparison to be entirely accurate (apples to apples)?

Good point. Looking at the equations and running the errors through, I estimate that if I get the circumference wrong by 1 cm, the V^2 term changes by about 1% which maps directly into my Cd term.
We really want to know distance per rotation, so I should take a run on a known course over 10 km say, record the RPMs vs time, integrate RPM and get meters/rev. Damn, another test! Of course my rho value is probably the biggest source of error.

Also, typically a coast-down test in one direction is usually sufficient - but since if you have to go out and back anyway, more data is usually better. I would suggest not averaging the run data, but instead comparing them independently.

Yeah, that makes sense. Maybe I should compile a table of results with several runs.

 

[DaveEV]

My only question on your methodology would be whether inflating the tire to higher PSI actually changes the effective rolling circumference enough to effect your calculation? I see only one assumed value of 2.05m being used. Did I miss something? Perhaps the elasticity of the tire results mostly in an increase in section width rather than circumference when PSI is raised, but if there is any actual increase in rolling diameter at higher inflation levels, wouldn't that have to be accounted for in your comparison to be entirely accurate (apples to apples)?

Good point. Looking at the equations and running the errors through, I estimate that if I get the circumference wrong by 1 cm, the V^2 term changes by about 1% which maps directly into my Cd term.
We really want to know distance per rotation, so I should take a run on a known course over 10 km say, record the RPMs vs time, integrate RPM and get meters/rev. Damn, another test! Of course my rho value is probably the biggest source of error.

A mark on the tire, tape measure and rolling the car 10 meters is probably sufficient if you can accurately measure to within a few mm.

 

[SanDust]

I don't see why I can't solve for Cd.

All you can measure directly is the power needed to overcome the drag force. But you don't know how much power goes to overcoming drag and how much is lost to drive train inefficiencies. You end up with equation P=1/2pv^3CdA. Since both P and Cd unknown you have two unknowns and one equation.

FWIW Nissan has said what it thinks Cd is. I can't remember but something like .29 or .31.

 

[sparky]

I don't see why I can't solve for Cd.

All you can measure directly is the power needed to overcome the drag force. But you don't know how much power goes to overcoming drag and how much is lost to drive train inefficiencies. You end up with equation P=1/2pv^3CdA. Since both P and Cd unknown you have two unknowns and one equation.

FWIW Nissan has said what it thinks Cd is. I can't remember but something like .29 or .31.

Yes, 0.29, I show that in my initial list of constants. I was merely solving for it to give myself confidence that my measurements and units were all kosher. Sorry if this seems dense but can you show me my formula error or error in my physical assumptions with the equation? : F = { Crr * M * g } + {Cd * A * 1/2 * rho * V^2} or a = { Crr * g } + { Cd * A * 1/2 * rho V^2 } / M
I'm directly measuring acceleration (at two Vs), not power. All your "drive train losses" are in Crr, no? Are you disputing that Crr is nearly constant with velocity? I'm confused.

 

[Smidge204]

Drivetrain losses would not entirely be linear. It would be a function of load, which itself is a function of speed and driving conditions.

But a few well-designed tests should be enough to characterize the vehicle apart from drivetrain losses. Costing tests will get you rolling resistance (which would include mechanical friction of drive components, and that's fine) and if you're brave you can try towing the LEAF and measuring the towing force at different speeds to get air+rolling resistance numbers. The only way to get pure air resistance numbers is to use a wind tunnel, which Nissan did to get the 0.29 Cd.

If you really want to isolate the wheels only, take them off the car and do coast-down tests by rolling just the wheels across a parking lot :lol:
=Smidge=

 

[SanDust]

All your "drive train losses" are in Crr, no? Are you disputing that Crr is nearly constant with velocity? I'm confused.

Drive train losses are entirely different and separate from RR. They are not constant. They're very high at low speeds, drop to their lowest level with moderate speeds, and then increase in more or less a linear fashion as speeds increase. They could be lower than RR losses or they might be three or four times or even five times greater than RR.

You'd need an efficiency map to have a good handle on it. My guess is that the Leaf's drivetrain is quite efficient, maybe 88%, at lower speeds but not very efficient, maybe only 70%, at speeds approaching its maximum.

 

[sparky]

All your "drive train losses" are in Crr, no? Are you disputing that Crr is nearly constant with velocity? I'm confused.

Drive train losses are entirely different and separate from RR. They are not constant. .....
...You'd need an efficiency map to have a good handle on it. My guess is that the Leaf's drivetrain is quite efficient, maybe 88%, at lower speeds but not very efficient, maybe only 70%, at speeds approaching its maximum.

Certainly, but you're describing electrical losses which I make no claim to be measuring with this test. I did calculate total energy loss from my calculated drag losses at 75 MPH over 100km, as an example. But, I understood that the drivetrain losses don't show up in that (and got a lower than realistic, kWh due to drag-only number of 17 kWh). I picked 85% efficiency for drivetrain out of thin air and got closer to 20kWh, which seems ok. Really, just another sanity check for my units etc.
That was not used to calculate Cd however.
Since I'm coasting and just measuring deceleration, those drivetrain losses are not part of the forces I see in this test. So, it seems to me that it is possible to calculate Cd the way I did. Granted, my accuracy may be poor due to non-linear nature of aerodynamics, my calculated air density, etc. But, nonetheless, it seems the equation I used, the way I used it, should yield Cd to some level (10%?).
Do you agree? If not, please show how that equation does not work for drag at a coarse level ( 0.33 > Cd > 0.26).

 

[tbleakne]

All your "drive train losses" are in Crr, no? Are you disputing that Crr is nearly constant with velocity? I'm confused.

Drive train losses are entirely different and separate from RR. They are not constant. .....
>>>

Certainly, but you're describing electrical losses which I make no claim to be measuring with this test. >>>
Since I'm coasting and just measuring deceleration, those drivetrain losses are not part of the forces I see in this test. >>>> But, nonetheless, it seems the equation I used, the way I used it, should yield Cd to some level (10%?).
Do you agree? If not, please show how that equation does not work for drag at a coarse level ( 0.33 > Cd > 0.26).

Sparky, I really like what you have done, as it relates closely to some measurements I have made.

Sparky's methodology looks perfectly fine to me. I would think it is Crr (and other things lumped with it) that is the important parameter to be measured because it is difficult to model. We have an accurate model for aerodynamic drag. My understanding is that solving for Cd was just a cross-check, since we do have the published value, which I am confident was determined in a wind-tunnel. We can solve for more than one parameter because we have data at different speeds, and the aerodynamic drag scales as V*V.

It seems clear that since the data is taken while coasting, electrical losses are not present. My interest is these results is to use them to subtract the mechanical losses out of the total loss under powered motion. Any remainder will be total electrical loss. I accept that these losses are low at low power levels. I want to see how significant electrical losses (battery, inverter, motor) are at higher power levels as a function of both speed and grade (altitude change). This will be useful in calibrating my ipod spreadsheet for predicting delta SOC for a particular trip.

Accounting separately for friction in the reduction box and wheel bearings might improve the measurement results, but only to the extent these losses scale differently than rolling resistance. It is reasonable to scale rolling resistance to total vehicle weight, as Sparky has done. I also believe it is common practice to treat rolling resistance drag as independent of speed. We could add a term proportional to speed but not weight to account for these other mechanical losses.

-----------------------

My own approach to taking data operates at constant speed and thereby avoids the numerical errors introduced by taking differences to measure derivatives. To use this method, you need a road that has a uniform negative grade.

1. You coast down this road in neutral (I use the zero power point because I am not comfortable with driving the car in neutral) and find the equilibrium terminal velocity. This is the point where all the drag forces (aerodynamic + mechanical) cancel the force of gravity projected onto the descending slope.

2. You drive back uphill on the same grade, keeping the same speed. The aerodynamic drag and other drag forces that are a function of speed remain the same. The gravity force has now reversed sign, but it still has the same magnitude as the drag, so the total power expended is twice the drag losses. You monitor the total power delivered from the battery over the course using the SOC meter (I have not completed its assembly). Call this Be.

3. You use Google Earth to measure the grade of the road you drove. This gives you the gravitational energy gain from your descent. Call this Ge. I claim that (Be - 2*Ge) is the total electrical loss.

4. You repeat the above steps for several different routes with different slopes to give a range of speeds, selecting routes with appropriate speed limits. I have found very gentle slopes of 2-3% give equilibrium speeds of 30 mph or so, and fairly steep freeway grades work at 60 mph.