*[Editor’s Note: This article appeared in the December 2013 issue. Some information may be different today.]*

Friction is what makes tires fun. The more they generate and the more grip they have, the more smiles they put on our faces. But that same friction that helps us cling to the road when clipping apexes also slows us down when we’re on our …

### The Formula:

We’re trying to figure out the force of drag (F). Since we know how much the car weighs (which gets us m) and we have the deceleration (a)–or at least the data to calculate it–we can solve the simple equation: F=ma.

Since we’re talking about testing the car on a relatively flat surface–or, in this case, averaging the data in opposite directions on the same surface–we can assume that the mass of the car acting against the drag is the weight of the car converted to mass.

To convert the car weight to mass, we need to divide the weight of the car, in pounds, by the gravitational acceleration. The gravitational acceleration for our purposes can be approximated at 32.2 ft./sec.2. [Ed. note: It varies with altitude; ask your local astronaut.]

This converts the weight of the car in pounds to the mass of the car in slugs. [Ed. note: Really, who comes up with these unit names?]

The acceleration is the difference in speed divided by the time between the two speeds.

The speeds from the start and the end of the deceleration need to be in ft./sec., not mph, so we’ll multiply our speed in mph by 1.466667.

The deceleration time needs to be in seconds and as accurate as you can get. Here’s the formula for deceleration: (starting speed–ending speed)/deceleration time.)

The deceleration is effectively a negative acceleration, so throw a negative sign in front of it, and the units are now ft./sec.2. For a relative number, you could divide the deceleration by 32.2 to get the g-load.

Now we have everything we need to calculate the drag force.

The output of F=ma will give a force in pounds.

This force is the average force on the car from your starting speed to your ending speed. Put another way, it’s the drag at the average speed.

This drag force is going to include everything from aero and tire resistance to bearing drag and any slopes in the road (which were eliminated by doing multiple passes in both directions).

Now we can convert that drag force in pounds to an equivalent horsepower. The basic formula is: horsepower=torque*rpm/5252.

Since we don’t have the torque or the wheel rpm, we need to calculate them.

We generated the drag for a given coast-down speed and need to convert it to torque at the drive wheels before we can convert it to horsepower. The only additional measurement we need to do this is the rolling radius of the tire.

Let’s take a wild swing and say the rolling diameter is 12.5 inches. That gets converted to feet by dividing by 12, giving us 1.04 feet.

The torque required to equal the drag is calculated by multiplying the drag force by the tire radius. This will be a pretty large number in relation to engine torque, but we have lots of gearing to amplify the engine torque as it makes its way to the ground.

Now that we have the torque at the ground, we need to use the speed conversion from the coast-down side to get the average speed in ft./sec. The average speed between 50 and 30 mph is 40 mph; converted to ft./sec., it’s 58.67 ft./sec. (multiply mph by 1.46667).

We need the circumference of the tire as well. That is the rolling radius multiplied by 6.28 (gives the diameter*pi). Do this for the value in feet so you don’t have to convert units.

Knowing the speed and the circumference, we can determine the rpm of the wheel. Since we know how fast the tire is moving along the road and how far one rotation of the tire will take us down the road, we can get the rpm by dividing the speed in ft./sec. by the circumference in feet and multiplying the whole thing by 60 (because there are 60 seconds in a minute; otherwise we’d wind up with revs/sec.).

Now we have all the numbers we need to get the horsepower (at the wheels) needed to overcome the drag on the car.

Following the formula, multiply the torque by the rpm of the wheel and divide the whole thing by 5252.

Out pops the horsepower, at the drive wheels, required to keep the car running at the average speed of the coast-down run.

### The Easier Option:

For those of you who don’t want to go back to school, we made things easier. If you go to grassrootsmotorsports.com/coastdown, you can download an Excel spreadsheet that does all the math for you. All you have to do is plug in the car weight, starting and ending speeds, time elapsed, and tire size. The spreadsheet will magically compute your own coast-down calculations. This is a great tool for determining not just rolling resistance, but aero drag as well, since the formula measures total drag in the system. You could even get crazy and use this formula to measure the effect of weight or driveline drag on a car’s performance.

Once we crunched our numbers for our TDI, we learned that, with our OE Continental tires, we were producing an average drag force of 88.25 pounds during our coast-down, or the equivalent of 10.6 horsepower. Our high-performance BFGoodrich street tires pushed that up to 98.51 pounds of drag, or the equivalent of 11.8 horsepower. That’s basically a 10-percent increase in tire drag, which made our 5-percent decrease in fuel economy completely believable.

Then we threw on a set of sticky BFGoodrich g-Force R1 race tires and repeated the test. The 245/40R18 R1s pushed our car’s total drag to a whopping 124.58 pounds of drag, or the equivalent of 15 horsepower. They do stick to the road, but obviously they’re costing us some straight-line speed.

For a little perspective, we took our Beetle to Virginia International Raceway to run it in the Ultimate Track Car Challenge and conduct some tire testing. We basically knew how the tires would finish on track, we wanted to see how much additional grip we were getting for our mileage tradeoff.

Although the finishing order was predictable, the results were still interesting. The OE Continentals were the slowest at an average of 2:39.1. They still provided a good bit of drivability, though. While they wouldn’t be our first choice for a track tire–that wasn’t what they were designed for–they held up admirably and provided good feedback regardless of their stated mission.

The BFGoodrich Super Sports were predictably faster by a full 2 seconds. Additional width and a more aggressive compound did their thing to improve lap times. In both cases of the street tires, though, lap times were held in check as much by the non-defeatable stability control as by the tires. While the intervention was not unsettling, we were definitely aware of its presence at times.

Fastest of all–not surprisingly–were the BFGoodrich R1s. They were more than 7 seconds faster than the Super Sports and nearly 10 seconds faster than the OE Continentals. Even more impressive, though, was the way the R1s seemed to work in concert with the stability control to turn it into an asset instead of a liability.

Our best guess is that the stability control is highly affected by slip angle and yaw rate, and the lower slip angles and yaw rates at which stickier tires naturally work are a much better companion to the digital overlord under the dash. There were more than a few times when we were heading through the uphill esses or into Hog Pen and getting ready to countersteer–to catch a slide or adjust the throttle to bring the car back on line–and the Beetle beat us to it. And it did so while losing an absolute minimum of momentum. Kudos to VW for designing a stability control that doesn’t punish us for having a little fun now and then. We’d still prefer an off button, of course.

So now you understand some of the tradeoffs a bit better between grip and grin. Those sticky tires will give you a great deal more grip in the corners, but they’ll also cost you power on the straights.