The 60-foot time: why the first 18 metres decide a run

On a drag strip the first timing beam sits 60 feet, or 18.29 metres, from the start. That is 4.5% of the 1,320 feet to the finish, so it looks like a minor line on the slip. It is the most useful one for judging a launch, because it is almost all traction and almost nothing else. This post turns a 60-foot time into an acceleration, then looks at why a loss there is so hard to recover. The time slip post covers the other lines.

From a 60-foot time to an acceleration

Treat the launch as constant acceleration from rest. It is an approximation, since real acceleration changes with every shift and every flicker of grip, but over 60 feet it is close enough to compare launches. Distance is d = ½ × a × t², so the average acceleration is a = 2d ÷ t². The speed at the beam, if acceleration really were constant, is twice the average speed: v = 2d ÷ t.

Take d = 60 ft and g = 32.17 ft/s². A 2.00 s launch gives a = 120 ÷ 4.00 = 30.0 ft/s², which is 0.93 g, and v = 120 ÷ 2.00 = 60 ft/s, about 41 mph. The other rows follow the same way:

The 2.44 s and 2.80 s rows are the grip-limited and wheelspinning launches worked out in the launch RPM post, where 2.44 s needs 0.625 g.

The sums also show how sharply the time reacts. Acceleration goes with 1 ÷ t², so a 60-foot time only 5% slower means about 9% less average acceleration: going from 2.0 s to 2.1 s drops it from 30.0 to 27.2 ft/s² (a factor of (2.0 ÷ 2.1)² = 0.907).

Why the loss is carried down the track

The 60-foot beam sits where the car is slowest, so it is where the car spends the most time per foot. At 41 mph it covers a foot in about 0.017 s; at the 120 mph of the time slip post's example it covers a foot in under 0.006 s. Time is recorded in the same units wherever it is lost, though. The clock does not run backwards.

The simplest case is a pure delay. Suppose two cars are identical in every way except that one reaches 60 feet 0.1 s later and then follows exactly the same speed-against-distance path. It will be 0.1 s behind at every later line. At a trap speed of about 172 ft/s (117 mph), 0.1 s is 17 feet, or about 5 metres, at the finish.

A pure delay is the best case for the slower car. Usually a weaker launch also leaves the car with less speed at 60 feet, and that deficit has to be made up by acceleration the car does not have to spare. To see the size of the effect, here is a deliberately crude model: after 60 feet, every car accelerates at a flat 10 ft/s² (0.31 g) for the remaining 1,260 feet. That is an assumption for the example. Real cars do not accelerate like that, and the ETs it gives are not realistic.

In this model, 0.1 s lost at 60 feet costs about 0.32 s at the finish, and the slower car also crosses the line a little slower. That overstates the effect, because it assumes the slower car then accelerates no harder than the faster one. A commonly cited rough rule of thumb puts it lower: each 0.1 s at 60 feet is worth roughly 0.15 to 0.2 s of quarter-mile ET. Treat that as approximate and dependent on the car. The pure delay and the crude model bracket it, and both say the loss at the finish is at least the 0.1 s lost at the start.

What decides the number

Over so short a distance, the 60-foot time is set mostly by how much force the tyres can pass to the road, not by peak power. That is why warm, clean tyres cut an example car's time from 2.80 s to 2.44 s with the same engine, and why too many revs at the start loses time rather than gaining it. Reaction time is not part of the 60-foot figure either: the clock starts when the car leaves the stage beam.

Because the early part of the track is so sensitive, it is also the part to change one thing at a time. Alter the launch RPM, run again, and compare the 60-foot line before looking at anything else. If it improves, the later lines start from a better position.