How this instrument works
The car leaves the ramp with velocity split into two parts: v·sinθ pointing up, v·cosθ pointing forward. From that instant only gravity acts on it. Setting the vertical position to −h, for a landing surface a height h below the ramp lip, turns the trajectory into ½gt² − v·sinθ·t − h = 0, a quadratic in t. That equation has two mathematical roots; only the one with the plus sign in front of the square root gives a positive time, which is why that particular root, not the textbook plus-or-minus, is what the formula uses.
Once time in the air is known, horizontal distance is a single multiplication: d = v·cosθ·t, because nothing pushes the car sideways once it leaves the ramp — gravity only pulls down. That is also the formula's honest limit: it treats the car as a point mass moving through a vacuum. Real aerodynamic drag and the pitching motion a ramp's kick imparts both trim a few percent off the distance this arithmetic predicts, so the number here is a planning figure, not a landing guarantee.
Stunt coordinators and ramp builders run exactly this pair of equations to decide where a landing ramp needs to sit before any car leaves the ground — misjudge the distance and the car either falls short of the landing ramp's up-slope or sails past it onto flat ground. A common carryover from a physics class is that 45° always maximizes range; that only holds when launch and landing sit at the same height. Once the landing surface is lower than the ramp lip, which it usually is on a real jump, the distance-maximizing angle is always below 45°, because the extra height a steep launch buys costs more forward distance than it returns.
- Set Launch speed to how fast the car crosses the ramp lip — the default is 45 mph, and the field also accepts km/h.
- Set Ramp angle, the ramp's slope above horizontal in degrees; steeper is not automatically farther once a height drop is involved.
- Set Ramp height above the landing surface to how far below the ramp lip the landing zone sits, in metres or feet; use 0 for a level landing.
- Read Time in the air, in seconds, for how long the car is off the ground.
- Read Horizontal jump distance to know where the landing ramp's near edge needs to be.
Worked example — 45 mph off a 15° ramp, 1 m drop
Take the calculator's own defaults: a launch speed of 45 mph (20.1168 m/s), a ramp angle of 15° (0.2617993878 rad), and a ramp height of 1 m above the landing surface. The vertical speed component is v·sinθ = 20.1168 × sin 15° = 5.206611 m/s. Feeding that into the time formula with g = 9.80665 m/s² gives t = (5.206611 + √(5.206611² + 2 × 9.80665 × 1)) ⁄ 9.80665 = 1.22793894077 s, which the instrument reports as 1.227939 s.
Horizontal distance then follows from d = v·cosθ·t = 20.1168 × cos 15° × 1.22793894077 = 23.8604949587 m, shown as 23.860495 m — about 78.3 ft, roughly two school buses nose to tail. The height field earns its keep here: drop the landing height to 0 m and the same speed and angle only carry the car 20.633226 m, nearly 3.2 m short of the 1 m-drop result, because there is less time for the ramp's forward speed to act before the car reaches the ground.
Questions
Why does the time-in-air formula include a square root?
Because the vertical motion is described by ½gt² − v·sinθ·t − h = 0, a quadratic equation in t, and solving a quadratic always brings in a square root. Two mathematical roots come out of it, but only the one with the plus sign gives a positive, physical time; the other root is a negative-time mirror image with no meaning for a car still sitting on the ramp.
What if the landing surface is higher than the ramp, not lower?
That case sits outside this field, which measures how far the landing sits below the ramp lip and bottoms out at 0 m for a level landing. Push h negative in the underlying equation and (v·sinθ)² can drop below 2g times the rise, leaving nothing real under the square root — physically, the car cannot reach a landing that high at that speed and angle.
Does a steeper ramp angle always make the jump go farther?
No. On level ground the classic optimum is 45°, but once the landing sits below the launch point, as height above the landing surface does here, the optimal angle drops below 45° and keeps falling as that drop grows. Past a certain angle, the extra seconds bought by launching more steeply upward return less forward distance than a flatter, faster-traveling path would.
Why doesn't the calculator account for aerodynamic drag?
Because it models the car as a point mass in free fall after leaving the ramp, the same simplification used for artillery range tables and ski-jump hill design. Drag depends on the vehicle's shape, speed, and attitude in the air, none of which ramp geometry alone predicts, so this figure is a starting estimate that real stunt planning refines with test jumps.
Can this formula be used for a motorcycle or bicycle jump instead of a car?
Yes. Projectile motion under gravity does not depend on mass, so the same two equations apply to any vehicle leaving a ramp at a known speed and angle — motorcycle stunt jumps and BMX ramps use the identical t and d formulas; only the launch speed and ramp geometry change.
Why is the ramp angle entered in degrees if the formula uses θ without a unit?
θ in the formula is in radians, the unit sine and cosine are defined against, but the Ramp angle field takes degrees because that is how ramps are measured and specified on site. The instrument converts before running the trigonometry — a 15° ramp becomes 0.2617993878 rad internally, so the conversion never has to be done by hand.