Stopping Sight Distance

Also known as SSD · braking distance for road design

d=vtr+v22ad = v\,t_r + \frac{v^{2}}{2a}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Stopping sight distance is two problems glued together: the driver notices nothing for the first two and a half seconds, then brakes. The reaction term is linear in speed, the braking term quadratic, which is why doubling the speed roughly triples the stopping distance rather than doubling it. AASHTO settled on a 2.5 s perception-reaction time — generous compared with the 1.0 to 1.5 s of an alert test driver, deliberately so — and a deceleration of 3.4 m/s² (11.2 ft/s²), a rate that most drivers can achieve on wet pavement without losing steering control.

A worked example at the American design values: 60 mph is 88 ft/s, so the reaction distance is 88 × 2.5 = 220 ft and the braking distance is 88²/(2 × 11.2) = 345.7 ft, totalling 566 ft — which is exactly the figure the Green Book tabulates for 60 mph. The formula as written assumes level ground; on a grade, the effective deceleration becomes a ± g·G/100, so a 6 % downgrade stretches that 566 ft to well over 600 ft. Enter the reduced deceleration directly if you need the graded case.

Stopping Sight Distance
d=vtr+v22ad = v\,t_r + \frac{v^{2}}{2a}
Where
  • dd= Stopping sight distance
  • vv= Design speed
  • trt_r= Perception-reaction time
  • aa= Deceleration rate