Ok guys I'm an aspiring game designer and I'm trying to develop a racing board game using real life as a basis for my mechanics. The problem im having is I'm not very mathematically inclined I found a formula on an urban planning website but I don't really understand it. Can anyone help me? enter image description here


2 Answers 2


The clearest way to make sense of any equation is to figure out the units. In this case, it's a bit ambiguous, but you know that velocity is m/h and that radius would be some kind of distance.

v^2 is a good assumption in this case, but it is not immediately obvious why without knowing something about the units of the denominator. Just by looking at this, it looks kind of like the well-known equation a = (v^2)/r which gives the acceleration a of an object traveling at velocity v in a circle of radius r. If you solve that equation for r, you get r = (v^2)/a, so it's a good guess that the denominator is in units of acceleration (if this is a valid formula).

If we are right so far, you can check that the units work out because we'd have v^2 in units of (miles^2)/(hours^2) and a in units of miles/(hours^2), which would give units of miles when divided. This is a unit of distance, so that's good. Let's keep working with that assumption for now and figure out what E and F are, and what all those magic numbers in the denominator are.

The basic assumption that this equation makes is that the car is travelling along the curve in the arc of a perfect circle, so all the acceleration is directly inward (no acceleration along the trajectory of the car). This makes the forces (accelerations) at work very simple -- you have the inward acceleration of the car and all other forces must cancel. So what force would cause the car to turn in a circle? The answer is friction!

When you turn a car's wheels, you are directing the force of friction to propel the car in a direction that is not straight ahead. If you are turning in a perfect circle, then any component of this friction force that is parallel to the motion of the car is canceled by some drag force(s) that are anti-parallel to the motion of the car, but there is no appreciable outward force to cancel the inward component of the friction force, and you get circular motion. The important thing to note here is that your formula has a "friction factor" -- which is actually just your friction force (you can look up the kinetic friction of tires on the road, but it's usually anywhere from 0.4 to 0.7 for wet to dry roads). In the end, you just need to note that the formula in question is using friction as the acceleration.

That's all good, but what about E and all those other random numbers? E is a factor that gives a ratio for banking of the curve. The more a curve is banked, the tighter the turn can be. Why is this? Because banking the road gives some component of the normal force (the force that is always perpendicular to the surface an object is resting on) towards the inside. In our flat situation, the normal force is perfectly canceled by gravity pulling down on the car. In the banked situation, gravity will only cancel the part of the normal force that is pointing straight up, and so the rest turns into part of our acceleration number.

Okay, that explains all the variables, but what about those numbers? Those numbers are why the units were ambiguous and we had to make assumptions earlier -- those numbers provide the units for us. I took the liberty and found the source for your image (corner radii) where they have helpfully labeled the units for us! R is given in feet, and V (as you noted) is miles per hour. E and F are unitless, as we expected. Since the denominator has to be one set of units, we can assume that all of the units are contained in the 15, and the extra 0.01 multiplying E is simply a scaling factor. That means that 15 has units of (get ready for this) squared miles per squared hour per foot, or more clearly, (miles^2)/((hours^2)*feet). How do I know this? Because the numerator ((miles^2)/(hours^2)) must cancel with the denominator to produce the left side, or feet.

So does 15 make sense? Well the force of friction is determined directly by the force of gravity, which in feet/s is about 32.2. We also know that to convert miles/hour to feet/s we have a factor of 1.47, so to convert the square values, we must square that, so we have 2.16. 32.2/2.16 = 14.9, so what that 15 means is actually just the acceleration due to gravity, so it does make sense!

The extra 0.01 just scales E because E is actually given as a percent (source), and must be multiplied by 0.01 to get a real ratio. It is important to note that both E and F come from the normal force acting on the car. E represents the direct normal force pointing inward (because the road is banked) and F represents the normal force converted to friction by the tires (and is just the ratio of kinetic friction between the tires and the road).


Edit: Someone pointed out that R is in feet. Makes sense as well, and the rest stands. When in feet, the formula probably describes a tightest recommended curvature instead of a comfortable curvature. If solving from the formula the speed for this scandinavian location, it would be 30 mph or 48 km/h. The speed limit there is 40 km/h, so 48 is uncomfortable.

Looks valid enough. An approximation, to provide a practical formula for solving the radius of a circle or arc - a segment of a road, and it's curvature.

If you look at this picture from a random american location you can see two measurements, both being ca 100 meters, and both being the radius of the road arc. When looking with Google, this curvature in fact seems quite common.

According to your formula (where it must be V^2, not V*2)

radius = (V^2) / (15 * (0.01 * E + F))

and assuming

  • the speed limit at that road part is 40 mph
  • the friction between the tyres and surface is almost 1 (we say 1 - you must check it)
  • we can ignore E

we get (40^2) / (15 * 1) which gives us 107.

Hence the conclusion would that the answer is in meters, and pretty close to the measured ca 100 meters radius (the google distance box says 196.55 m, which is the sum of both line segments, hence each of them is ca 100 meters).

You'd have to pick a number of similar cases, and see how well the formula fits reality (and ofc also check that it really is meters it outputs).

  • Higher speed: Bigger radius
  • Higher friction: Smaller radius
  • Lower friction (eg. gravel): Bigger radius

Makes sense.

  • \$\begingroup\$ This answer is actually wrong, the formula is lifted from nacto.org/docs/usdg/… which clearly denotes the radius as feet -- a 100 m radius is GIGANTIC, and would represent a very shallow curve. \$\endgroup\$
    – dannuic
    Commented May 2, 2016 at 19:07
  • \$\begingroup\$ It's a 1000-page document, you could maybe point at a page number? As said, the unit (and friction c) needed to be verified in any case. I'll edit and insert a note. Thx. \$\endgroup\$
    – Stormwind
    Commented May 2, 2016 at 20:53
  • \$\begingroup\$ Certainly, I didn't think of including a page number because I just searched the document for the word "radius" -- The section you want to read is "HORIZONTAL ALIGNMENT" starting on page 172 of the pdf (which is physical page number 131). The actual formula is in the "Minimum Radius" subsection starting on pdf page 183 (which is physical page number 142). \$\endgroup\$
    – dannuic
    Commented May 3, 2016 at 14:38
  • \$\begingroup\$ Thanks. And there they are, metric and US Customary, side by side :-). It's an impressive document alltogether. \$\endgroup\$
    – Stormwind
    Commented May 3, 2016 at 16:02

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