# Forums

## Here's An Engineering Problem For You

Everyone knows that excessive acceleration accompanied by corresponding braking is an inefficient way to use stored energy, whether it be for ICE or EV's. However, what about the situation where you accelerate from one speed to another at a constant rate, and hold the final speed (like pulling onto a highway)? I contend that it doesn't use any more energy whether you creep up to speed or make it fun.

Here's my reasoning: the work done to accelerate a rigid body just depends on the initial and final velocities, not the acceleration rate. If you accelerate twice as fast it takes half the time to reach your final speed, and you use the same amount of energy.

Now cars are not rigid bodies, and energy is lost due to friction, heat and the inefficiencies of the conversion of stored chemical energy to mechanical power. However, if these losses are linearly related to the acceleration rate, then again the energy used to accelerate from one speed to another is the same no matter how fast you accelerate.

Now here's my final curve: the aerodynamic drag might be negligible for the few seconds of going from 30 to 65 mph, but it becomes more significant the longer it takes to do that. So if you accelerated at a very low rate, and it took you an hour (or many hours) to go from 30 to 65 mph, then the aerodynamic drag will become a more significant factor in the energy used, and you will use MORE energy than if you accelerated faster.

So I contend that putting your foot into it to pull onto a highway (as long as you don't exceed your desired final speed) doesn't use any more kW-Hrs than poking along. So (safely) enjoy yourself! Would be great to hear from a Tesla engineer on this one!

Volker.Berlin | October 24, 2012

Can someone please find the thread where this has been discussed at length?

Timo | October 24, 2012

Now here's my final curve: the aerodynamic drag might be negligible for the few seconds of going from 30 to 65 mph, but it becomes more significant the longer it takes to do that. So if you accelerated at a very low rate, and it took you an hour (or many hours) to go from 30 to 65 mph, then the aerodynamic drag will become a more significant factor in the energy used, and you will use MORE energy than if you accelerated faster.

Only if you time the time used in acceleration. If you just time the same time to both cars, or the distance, then the one that accelerated faster use more time in higher air drag speeds, which means it uses more energy.

Whity Whiteman | October 24, 2012

I had a friend at school and in the age of 11 he had a good idea: If You go double fast, You need half the time... so you need less gasoline, or at least the same amount.
That's -for sure- nonsense.
From 0-60 in 4 Seconds uses much more energy then from 0-60 in 20 seconds...

Timo | October 24, 2012

Assuming without losses 0-60 in 4s and same in 20s you end up with same kinetic energy for both cars, so there should not be any difference between the two. Difference comes from the losses in efficiency and time used in higher loss speeds. For BEV those two should be a lot closer to each other than with ICE because of very high efficiency of the motor regardless of accelerator pedal position.

Mark K | October 24, 2012

Volker's right, discussed at length before. Summary:

Serious calc's require accurately modeling many variables to honestly quantify this. But here are some topline observations:

1. In ideal (unreal) conditions ... Perfect battery impedance, no drag, etc., yes, the energy it takes to accelerate the car is the same whether done fast or slow.

But in the real world, the losses are significant. Battery electrochemistry results in higher, nonlinear thermal loss at very high current (i.e. rapid acceleration). And as Timo points out, the time/miles spent at the higher aerodynamic drag speed will affect the budget too.

2. Aerodynamic drag dominates above 35 mph. (so it's the major loss on freeways, even at steady state speed without acceleration).

3. Despite the losses from nonidealities, an EV is far less lossy on any given acceleration profile than an ICE.

So in that sense, yes, live large and gun it (relatively) guilt-free.

Mark K | October 24, 2012

Further to that, aerodynamic drag increases as the square of velocity so think of it as a miles thing rather than time thing.

If you drive 10 miles at 30mph, your drag loss per mile will be just 1/4 the loss at 60mph.

Regardless of the longer journey time, the actual total energy loss due to drag is 4 times greater on the same trip distance.

So aero drag is, uh, a real drag.

That's why TM put so much engineering effort into getting the lowest Cd figure on the market.

Volker.Berlin | October 24, 2012
Timo | October 24, 2012

I wonder if you could actually save energy with rapid acceleration to about 20-25mph, and then slow the acceleration down. That's the sweet spot for Model S, at 20mph your range is over 450miles with 85kWh battery.

Brian H | October 24, 2012

Heh. "rapid acceleration to 20mph" sounds amusing. How would you distinguish it from slow acceleration? ;)

Getting Amped Again | October 24, 2012

Timo - Sorry I don't follow this:
"Only if you time the time used in acceleration. If you just time the same time to both cars, or the distance, then the one that accelerated faster use more time in higher air drag speeds, which means it uses more energy."

"Both" cars are going the same speed at the beginning and the end. The one that accelerates slower has more losses due to aerodynamic drag than the one the accelerates faster because it takes longer to achieve that final speed. The aerodynamic drag is most-likely insignificant compared to the energy required to accelerate the vehicle however - it's more important at constant highway speeds, where you are just overcoming friction, drag and chemical-to-mechanical inefficiences.

Whity Whiteman "That's -for sure- nonsense.
From 0-60 in 4 Seconds uses much more energy then from 0-60 in 20 seconds..."

Actually Whity I would be 100% correct if all the losses where linearly related to the acceleration rate. Mark K points out that they are not, and that's the only hole in my hypothesis.

It would be interesting to know how nonlinear the losses are. I would bet that most people (like Whity) would assume that 2X acceleration uses 2X more energy, and I would also bet it's no where near that - probably 10-20% more. Anyone know the facts on this?

Carl Barlev | October 24, 2012

@Getting Amped Soon

You seem to think drag is proportional to acceleration rate. You suggest that drag losses will be greater if you spend more time accelerating between two speeds.

I think this is where your logic has taken a wrong turn. Drag losses are not proportional to rate of acceleration, but rather to car speed at each point along the acceleration curve (roughly speaking). Higher speed = more drag.

A simple example:

Car A takes 1000 m to reach final speed.

Car B reaches final speed in 500 m and then maintains final speed for the next 500 m (same total distance as Car A).

Car B has a higher average speed over the 1000 m and will thus experience higher drag losses over this distance.

Getting Amped Again | October 24, 2012

Here are some more facts (did the calculations in a hurry, hope they're right):

Assuming it takes 3 seconds to accelerate from 30 to 65 mph, the constant force required on a 4647 lb weight is 2469 lbs.

Assuming a frontal area of 20 sq. ft., and a drag coefficient of 0.25, the drag force at 65 mph is 106 lbs. At 30 mph it's 23 lbs.

So the aerodynamic drag is only between 1-4% of the force required to accelerate the car in my scenario - not that significant.

zwjohnston7 | October 24, 2012

Also, your electrical losses increase at I^2 (current squared). This is defineatly not linear. The faster you accellerate the more current is used at once and you have much more power lost to heat in all the wires. That heat by the way also increases the resistance of the wires which makes your losses worse yet.

Other losses include air drag (as mentioned) and friction (tires on the road and all the moving parts of the car) and NONE of these are linear in nature.

All the losses do add up, and that is why you are hearing about how "spirited" driving and drag racing with your Model S can drastically reduce your range.

The only thing I agree with you on is that the efficiency of doing this in an EV is better than doing it in an ICE.

GoTeslaChicago | October 24, 2012

you're making this more complicated than necessary.

Although there is no acceleration penalty in the model S like there is in an ICE car, slower acceleration will still result is less energy consumption simply because the car spends more time at the slower speeds where aerodynamic drag is less.

Let's perform a thought experiment.

Suppose your choice is 0 to 60 in 4.4 seconds and then drive at 60 mph for one hour, or take an entire hour to go from 0 to 60. In the first case your energy consumption is basically that of a car going 60 mph plus 4.4 seconds at maybe 6(?) times that rate.

In the second case you energy consumption will be close to that of a car averaging 30 mph.

However, if you're going to spend an hour driving at freeway speeds, the difference between getting there in 5 seconds, vs 30 seconds will have only a very tiny effect on the average energy consumption for the hour as a whole.

GoTeslaChicago | October 24, 2012

I probably should have added that the measured parameter is energy consumption per total travelled distance, not per time period.

Getting Amped Again | October 24, 2012

Can anyone quantify their opinions with engineering data or specs? That's what I'm really after and I'm only getting qualitative responses, and they range from I'm essentially right to I'm dead wrong.

DouglasR | October 24, 2012

I thought that when you stomp down on the accelerator, it squirts a whole lot of electrons into the carburetor, which then don't burn as efficiently.

Getting Amped Again | October 24, 2012

GoTeslaChicago - I'm only talking about the energy used to change speed from 30 to 65 mph, nothing after that. Aerodynamic drag is insignificant in my scenario - see my post and calculations explaining why.

nickjhowe | October 24, 2012

There are two different questions here I think. One is about total energy consumed over a period of time or over a distance under a combination of acceleration and constant speed. The other is about energy consumed at different accelerations. They are related but not the same.

If the question is: is it better to mash the throttle or take it easy (with easy being some level less than "mash") then the end point needs to be defined. A better question is "what is the lowest energy way of getting to xx mph" (a follow up question might be around constant or variable acceleration)

There are two different energy components - (A) using chemical energy to increase kinetic energy (acceleration) and (B) using chemical energy to maintain kinetic energy (speed). The former is most effected by efficiency of conversion. The latter is most affected by duration and drag (CD, rolling resistance). Neither measures are linear with time.

If we accelerate twice as hard, (A) may be higher or lower depending on the efficiency of energy conversion. If it takes twice as long to get from one speed to another, then the car will spend twice as long at each incremental speed, and therefore (B) will be some multiple as large.

So...if it is a 0-60 question, sum (A) and (B), integrating over time. Anyone want to throw some numbers into Matlab?

Thumper | October 24, 2012

If an apple a day keeps the doctor away, does two apples keep two doctors away for one day or one doctor away for two day?

Getting Amped Again | October 24, 2012

Nickjhowe - really I'm just after your second question. At a constant speed it's really all about how fast you are going. My question is about accelerating from one speed to the next, like pulling onto a highway. Most people would assume that accelerating twice as fast uses twice as much battery energy. I contend that it's no where near that but I'm looking for hard numbers.

zwjohnston7 | October 24, 2012

It definitely wouldn't use twice as much ENERGY to accelerate twice as fast. It would require twice as much FORCE (assuming no losses). In a lossless environment it would use the same amount of energy. The losses associated with accelerating twice as fast are not only twice as much, more like 4 times as much since electrical losses and air drag losses will tend to increase as a squared function. No one here is going to do real hard numbers for this cause you would need to know exactly what all of these losses are. Do you know the impedance of the motor and all wiring or the frictional losses of the gear and tires? What about losses associated with drawing power from the battery? Nope. Not to mention it will depend on the temperature outside and how long and how hard you have been driving prior to this one particular acceleration.
But a good estimate is probably accelerating less than half peddle to the floor probably uses 10-50% less energy for the acceleration period as opposed to flooring it.
When you get your car you can run a couple hundred field tests and record all the data and then you'll know. Let me know how close my wild guess is.

Tiebreaker | October 24, 2012

Great!!! What other scientific excuses can we find to justify the pure joy of insane acceleration?

Carl Barlev | October 24, 2012

Getting Amped Soon,

I am an electrical engineer. I majored in power systems and power electronics and also took papers in aerodynamics and mechanical systems.

Unless I have misunderstood the context of your original question, I can assure you that crunching hard numbers won't increase my certainty of the fact that accelerating faster will use MORE energy for any given trip.

As you correctly noted, the gain in kinetic energy will be the same regardless of how you get up to speed. The only difference then will be in which acceleration path has the higher losses.

There are three major loss factors and they all follow the same relationship, namely that faster acceleration will increase losses.
* Higher drag losses over acceleration distance
* Higher resistive losses in wiring
* Higher losses in battery

We don't need to know how much greater each of these will be, to know that the total losses will be greater when you accelerate at a faster rate.

Sorry Tiebreaker, but the science doesn't justify insane acceleration (if you goal is energy efficiency). Doesn't mean you can't enjoy though :)

evanstumpges | October 24, 2012

Another variable to throw into the mix is motor efficiency, which is not constant across the torque/speed range. Motors tend to be less efficient at low speed and low torque. I haven't seen an efficiency map for Tesla's AC Induction motor, so I can't quantify this or even know to what extent efficiency drops off for their motor at low speed/torque.

This said, I agree with others that faster acceleration results in greater overall losses in the real world.

Getting Amped Again | October 24, 2012

Carl Barlev - you're not correct about the drag losses. In the scenario I presented they are insignificant (see my previous post) plus their effect is more pronounced the longer the acceleration period. The resistive drag force is proportional to speed, not acceleration, and the speeds are the same in both scenarios.

As others have pointed out, it really only matters if the mechanical and electrical losses are nonlinearly related to the acceleration rate, and a few have proposed that indeed they are. I don't doubt that, but nobody can quantify it. Everyone just says "more" and "higher", just as you have. That's not what I'm after but thanks for your reply.

Tiebreaker | October 24, 2012

@Carl Barlev: I am an EE too, that was :-P . Coz that what the thread really does - looking for excuses... :-D

Tiebreaker | October 24, 2012

:-P -> tongue in cheek

Brian H | October 25, 2012

The "terminal" speed is the same, but the higher accel version spends more time at high speed. That's what high accel is for.

Timo | October 25, 2012

@Getting Amped Soon,

I did mean exactly what I said. If you time only the time used to accelerate then the faster acceleration rate gives you less loss, because you spend less time in higher loss speeds, but that car has also a) gone less distance compared to other car IE. has done less work before it reaches the same speed, and b) at that same timepoint the faster acceleration gets to the high speed the other car has used less energy because it has not yet got to the high loss speeds.

To really compare the two cars you need to give them either same distance or same timeframe where both cars have reached the same speed. In both of these cases the car that accelerates slower wins. You get twisted results because you stop the measurements in different points for both cars.

Carl Barlev | October 25, 2012

@Tiebreaker

Yes, I can see now that your comment was :-P (brain was still stuck in "serious" mode after my previous post and took your words at face value).

It took me a good few seconds to work out what :-P mean too and I only saw your definition below after I'd figured it out - guess it pays to look ahead :)

Carl Barlev | October 25, 2012

@Getting Amped Soon

I'm confused by what I see as contradictions in your original statements and contention, but it could also just be that I've misunderstood your intended meaning...?

You say that you are only interested in the energy needed to accelerate to your final speed, but the purpose of your question seems to be to understand how this will affect overall energy efficiency.

This is two different questions:
1) Energy to accelerate to final speed, versus
2) Energy to accelerate over fixed distance

If you want to know how different acceleration rates will affect the efficiency of your energy usage when you "pull onto a highway", then you must measure the energy losses over a fixed distance.

From the fact that you want to disregard the constant-speed component of the total losses, it seems that you are considering the first question...

Getting Amped Again | October 25, 2012

Carl Barlev - yes I'm only interested in the acceleration portion of the problem (your #1). Battery usage at constant speed is well understood: it's rolling friction and aerodynamic drag, and I think the aero drag dominates and increases with the square of the speed.

I confused things in my original post with my statement about extremely slow acceleration rates as my dumbed-down "limit analysis". Should have just left that out.

This all stems from my frustration when I get behind a slow-poke when pulling onto a highway or away from a stoplight. They think they are saving gas (or in this case charge), but what really hurts your gas mileage is excessive acceleration and braking - like zooming up to a stoplight and jamming the brakes and always driving like that.

People are brainwashed into thinking that puttering onto the highway is saving them gas, but I contend that the savings is insignificant. As us engineer-types understand, if the system losses are linearly related to the acceleration rate, it makes NO difference what the acceleration rate is. If you want to accelerate at 1.0X and you have 10% losses, you have to use 1.1X of power. If you want to go 2.0X you have to use 2.2X of power, but you use the higher power for 1/2 the time, and the total energy consumed is the same in both cases.

Some posters have pointed out that the electrical losses are nonlinear, which I totally get. (Some posters have also talked about aerodynamic drag - not significant in my problem.) I was just hoping that someone had some real data on just how nonlinear the electrical losses (or inefficiencies) were. If I had to guess, there is a sweet-spot that the motor operates at, and it's somewhere between creeping onto the highway and flooring it. I also think that "pretty decent" acceleration is not any more wasteful than puttering, and might even be less so due to the fact that the motor might not be very efficient at low RPMs or loads (I don't know anything about electric motors).

It's aggressive stop and go driving with excessive braking, and driving at high highway speeds that kill your range. I contend that in situations where you want to increase your speed to some "cruise" level, it doesn't make that much difference if you poke up to speed to go 3/4 throttle to get there. I was hoping for some facts to prove or disprove my theory (as I knew I had rigid-body physics on my side).

Some engineer at TM has all this built into a spreadsheet and could answer my question in two minutes. I was hoping he or she might read my post.

Thanks.

GoTeslaChicago | October 25, 2012

Getting Amped Soon

"Some posters have also talked about aerodynamic drag - not significant in my problem."

Not sure why you keep saying that aerodynamic drag is not significant. You haven't documented that, other than to say that it is well known. However, it is the one factor that is definitely known to be non- linear.

kalikgod | October 25, 2012

I am too tired tonight, but I will try to put a reasonable aerodynamic study of this together this weekend.

Getting Amped Again | October 25, 2012

GoTeslaChicago and kalikgod - I'm only talking about accelerating from 30 mph to 65 mph, and nothing more. I agree my original post went off into "very slow acceleration" as an FYI, but I'm just talking about the acceleration portion. Aerodynamic drag is miniscule compared to the force to accelerate a 4647 lb weight! It comes into play only when traveling at a constant speed, and all you're overcoming is friction and drag.

Here's my previous post, which you missed. Feel free to check my calculations.

Here are some more facts (did the calculations in a hurry, hope they're right):

Assuming it takes 3 seconds to accelerate from 30 to 65 mph, the constant force required on a 4647 lb weight is 2469 lbs.

Assuming a frontal area of 20 sq. ft., and a drag coefficient of 0.25, the drag force at 65 mph is 106 lbs. At 30 mph it's 23 lbs.

So the aerodynamic drag is only between 1-4% of the force required to accelerate the car in my scenario - not that significant.

jbunn | October 25, 2012

At a guess, I would say accelerating smoothly in an ICE IS more fuel efficient, but I think it's because the ICE has better efficiency in a temperature range and an RPM range. Don't have facts to back this up, but I think power varies in an ICE based RPM per unit of fuel. I think that might explain the hesitation of hypermilers to floor it.

Timo | October 26, 2012

@Getting Amped Soon, you seem to have mixed statements.

This one People are brainwashed into thinking that puttering onto the highway is saving them gas, but I contend that the savings is insignificant.

Talks about gas savings in general, and people saying that are quite correct, it saves them fuel.

however this one yes I'm only interested in the acceleration portion of the problem

is talking about half of the problem, not the whole picture. You are comparing different results using different endpoints, which gives you twisted results.

Energy you put into car is independent of the acceleration rate, you are correct about that (you end up with same kinetic energy, but then the difference comes from that point forward where your faster car has already accelerated to end speed (which is the point where you stopped your calcs) and is going in constant speed, while the other car is still accelerating. That gives you another set of energy to consider: how much energy you use to travel the same distance. That's the one where the savings in fuel comes from.

Slower acceleration gives you better result because you spend less time in higher loss speeds. It's not big difference if both cars accelerate in reasonable rates but it is there.

Comparing two different sets of losses using same endpoint in time or in distance (instead of speed) is kinda nasty integral and I'm way too tired to make calculations in my head for that.

Carl Barlev | October 26, 2012

+1 Timo

@Getting Amped Soon,

I see now why you were saying that drag is insignificant. You were refering to the fact that drag losses will be very small compared to the total energy expended during acceleration.

This is correct, but this doesn't mean that you can disclude drag losses in an energy efficiency comparison of different acceleration rates. You must compare apples with apples... or in this case, energy losses over any fixed distance (must be the same in each case).

Overall, you will use less energy if you accelerate slowly and steadily. Perhaps not during the actual acceleration, but certainly once you factor in the additional losses from traveling further at a higher-loss speed.

Getting Amped Again | October 26, 2012

Timo and Carl Barlev - I'm only talking about the energy used to accelerate from one speed to another - an event that lasts 3-4 seconds. I confused the issue with next-to-last paragraph, which I wish I could edit out. I was just using an extreme example of a very slow acceleration, and it took the post off into a different direction and more complicated problem.

Focusing on the acceleration problem alone: most people think that slowly accelerating to highway speed saves them energy, and 2X acceleration uses 2X more energy (gas or electrical). In fact it's not anywhere close to that. If losses were linear, it wouldn't matter what your acceleration rate was. I was hoping someone could quantify the battery and motor behavior throughout its torque range with real data - like "3/4 accel uses only 10% more energy than 1/4 accel when the change in speed is the same in both cases." I don't think I'm going to get that information from this post, but someone at TM has it in a spreadsheet.

Work (Energy) = integral of Force over a distance

Let's ignore rolling resistance. At any speed the force required merely to overcome drag and maintain velocity increases cubically with velocity. Accelerating force is applied on top of that and is greater for a car accelerating faster.

Since s = (v+u)t/2 and t is inversely proportional to the constant acceleration, distance traveled is inversely proportional to the rate of acceleration.

So, accelerating from a stop, just overcoming drag, a car accelerating twice as fast will reach target speed in the first half of the distance of the car accelerating more slowly while having to apply forces to overcome drag that are at least 8 times as great at each point.

Assuming a perfect motor I'd suggest it'd need to do 4 times the work to overcome the drag and then using the constant F=ma the acceleration work would be the same.

Of course there's rolling resistance to overcome, so you could conceivably use more energy for a given _period of acceleration_ but because, unlike ICEs a simple electric drivetrain has a faster=worse efficiency rule, for a given _distance_ slow acceleration is better. I don't know how rolling resistance works but I'd guess that's why traveling really, really slowly is inefficient too.

Oh damn, there's no delete. They'll each be traveling the same speeds at some point, so the slow car will end up doing double the drag work?

But it's more efficient to accelerate slowly because to cover the same distance the fast car will be traveling at a higher average speed so with higher average drag.

nickjhowe | October 26, 2012

OK - here we go. I've dusted off my calculus and noodled on this for a while. If you want to see the calculations check out this link. can't guarantee they are correct, but the best I can do.

Imagine three scenarios:
Scenario a: accelerate from 35-60mph in 2s;
Scenario b: accelerate from 35-60mph in 10s;
Scenario v: accelerate from 35-60mph in 2s then drive at 60mph for 8s.

Assuming motor efficiency/losses are constant regardless of load (!!), then

Energy (a) = 518kJ; Energy (b) = 594kJ
So slow acceleration takes 14.5% more energy (because it takes longer!)

Energy (c) is 628kJ (5.6% more than (b))
So if you are accelerating onto a freeway, you will use less energy over the same time if you accelerate slower (but you'll also get there a little later)

FYI In scenario (a), 96% of energy is acceleration; 4% is overcoming drag/rolling resistance. In Scenario (b) it is 86/14 (but the absolute energy to accelerate is the same)

If motor losses vary by more than 5% under heavy load then the calc may go the other way.

Getting Amped Again | October 26, 2012

Thanks nickjhowe! I knew I was right of course, but people will point out that the motor losses are nonlinear with current/rpm/load. How nonlinear they are is really the crux of the problem.

Your scenario C is interesting and correct, but misleading. Trips are based on distance, not time. Can you re-run for trips of varying distance, say 1, 5 and 10 miles? You don't have anything better to do, right?

nickjhowe | October 26, 2012

At 60mph, aero drag uses about 322kJ per mile, rolling resistance about 499kJ per mile for a total of 821.

It takes about 750kJ to go 0-60 in pure kinetic energy. With fast acceleration, drag is small (25kJ) as the car only covers 60m.

So - 0-60 in 4.4 is about 1 mile of driving at 60mph, give or take. (at least according to my possibly flawed calculations)

nickjhowe | October 26, 2012

Interestingly, with such low Cd, it is only above 60mph that aero drag is more than rolling resistance (based on a coefficient of resistance of 0.015 for the tyres). If anyone has a better estimate I can update the graph:

(y Axis is in N; x axis is MPH)

Brian H | October 26, 2012

That aero stuff is really a drag!

Hence the appeal of vacuum tunnels ... ;)

Getting Amped Again | October 26, 2012

nickjhowe - great work! I have one more request:

on-ramp acceleration from 30-65 in 3.0 seconds, total trip distance 1.0 miles.
Same as above but the on-ramp accel period is 6.0 seconds.

I think you are saying that, as your trip gets longer than a few miles, what you did to get on the highway becomes less important with respect to the range used up. I'm interested if it matters in a one mile trip.

Also - pulling away from a stoplight in a one-mile trip, same accelerations scenarios as above (except 0-35 mph). Drag is much less significant in this case (as you know).

I think these two scenarios are going to show that puttering around in a Model S doesn't improve your range, at least when accelerating to a "cruise" speed.

Thanks man!

Getting Amped Again | October 26, 2012

I meant "what you did on the on-ramp becomes less important..."

nickjhowe | October 26, 2012

Too much fun at a Haloween party tonight, I'll cruch the numbers tomorrow. :-)