Car Jump Distance Calculator

Created by Dominik Czernia, PhD, Álvaro Díez and Miłosz Panfil, PhD
Reviewed by Steven Wooding and Jack Bowater
Last updated: Jan 09, 2023

The car jump distance calculator is here to help you simulate the very dangerous but, for sure, phenomenal stunt – jumping in the car from one ramp to another. Remember that we should consider many different factors during a car jump, but we can't take them all into account.

Even if it isn't 100% accurate, the car jump distance calculator still allows you to estimate the expected jumping range and landing angle including the air drag force. These are crucial parameters and should be known before any trial jumps.

Keep reading to learn more about the calculator and physics behind the car jumping problem!

How to use the car jump distance calculator?

We divided the calculator into three parts.

Basic jump parameters

That's the central part which is obligatory to see any results. You need to provide:

  • Take-off ramp slope and height – the slope should be expressed in angle units;
  • Landing ramp height – note that we show the results for only first part of the jump if the landing ramp is too high for a jumping car;
  • Take-off speed – the speed that the car has just before take-off; and
  • Standard gravity – by default, we set it to 1 g1 \ \mathrm{g}, which is the mean gravitational force on Earth. You can adjust it to the standard gravity at your location or change to standard gravity of other planets.

That's enough to analyze a simple projectile motion problem. For more advanced computations, you need to input more values. Decide whether you want to include the air drag force, compute car tilt during a jump, or both. A new section called "Advanced calculations" should appear.

Advanced calculations

Regardless of what advanced options you choose, you need to specify car mass, length, and height. The former is the total mass of a car including all equipment, fuel, and the passengers' weight. Additionally, the air drag force option requires:

  • Air density – which for standard conditions is 1.225 kg/m31.225 \ \mathrm{kg/m^3}. Use our air density calculator to find the density of air for your specific jumping conditions; and
  • Automobile drag coefficient – you can check the drag coefficient for various cars, e.g., here.

The car tilt computation needs the following additional parameters:

  • Wheelbase – the distance between car's rear and front axles;
  • Horizontal and vertical locations of the center of mass – in other words, the longitudinal and altitudinal positions (see the picture that should appear in the calculator), which you can estimate using our car center of mass calculator;
  • Car height – the car height is required to approximate the moment of inertia (if you don't know its value) and air resistance; and
  • Moment of inertia – if you know the exact value of the moment of inertia, you can input it directly in the corresponding field. This moment of inertia is around the axis passing through the car's center of mass and parallel to the car's width (the calculator will move the axis of rotation to the correct place automatically). Read the FAQ below to learn how we find the approximation of mass moment of inertia in the car jump distance calculator.
Four solid cuboids as car's moment of inertia approximation.
A car is modeled as four solid cuboids to calculate the moment of inertia displaced by a certain distance from the center of mass. See FAQ below for more details.

Results

The results section consists of:

  • Summary table – includes the most interesting car jumping parameters like range, time of flight or landing tilt angle; and
  • Customized chart – specify what should be on the x-axis (horizontal position or time) and y-axis (position, velocity, acceleration, or car tilt angle). You can add up to three graphs at the same time.

Click the advanced mode button to change the units of quantities appearing in the table or chart.

Determination of the distance traveled mid-air

You can treat the launching of a car from a ramp as a projectile motion with an initial velocity v0v_0 and angle of launch α\alpha. The standard way to analyze a projectile motion trajectory is to treat the horizontal xx and vertical yy positions as separate variables. We'll use differential equations, which is a suitable method to deal with more complex problems (e.g., including drag force). If you want a simpler approach, visit our projectile motion calculator.

Car jumping initial parameters with a ramp and velocity decomposition.

The only force acting upon the car is the gravitational force, equal to the mass mm multiplied by the standard gravity gg. Furthermore, we can decompose the velocity vector into two perpendicular vectors using the sine and cosine trigonometric functions. The first component is in the horizontal direction v0x=v0cosαv_{0\mathrm{x}} = v_0 \cos \alpha and the second one in the vertical direction v0y=v0sinαv_{0\mathrm{y}} = v_0 \sin \alpha. The two sets of differential equations are then:

2yt2+g=02xt2=0\frac{\partial ^2 y}{\partial t ^2} + g = 0 \\[1.5em] \frac{\partial ^2 x}{\partial t ^2} = 0

with initial conditions:

x(0)=0y(0)=h0xtt=0=v0xytt=0=v0yx(0) = 0 \\[1.5em] y(0) = h_0 \\[1.5em] \frac{\partial x}{\partial t}_{|t = 0} = v_{0\mathrm{x}}\\[1.5em] \frac{\partial y}{\partial t}_{|t = 0} = v_{0\mathrm{y}}

where h0h_0 stands for initial height and tt for time since take-off. We'll skip the whole procedure, but it shouldn't be hard to find the solution for a person who is familiar with differential equations:

x(t)=v0xty(t)=h0+v0ytgt22x(t) = v_{0\mathrm{x}} t \\[1.5em] y(t) = h_0 + v_{0\mathrm{y}} t - \frac{g t^2}{2}

Just for the sake of completeness, we listed some of the most interesting parameters that you can derive from the above equations.

Velocities

vx(t)=v0xvy(t)=v0ygtv_{\mathrm{x}}(t) = v_{0\mathrm{x}} \\[1.5em] v_{\mathrm{y}}(t) = v_{0\mathrm{y}} - gt

Accelerations

ax(t)=0ay(t)=ga_{\mathrm{x}}(t) = 0 \\[1.5em] a_{\mathrm{y}}(t) = -g

Maximum height hmaxh_{\max} and associated time tmaxt_{\max}

hmax=h0+v0y22gtmax=v0ygh_{\max} = h_0 + \frac{v_{0\mathrm{y}}^2}{2g} \\[1.5em] t_{\max} = \frac{v_{0\mathrm{y}}}{g}

Time of flight tft_\mathrm{f} and range x(tf)x(t_\mathrm{f})

tf=v0y+v0y2+2g(h0hf)gxf=v0xtft_\mathrm{f} = \frac{v_{0\mathrm{y}} + \sqrt{v_{0\mathrm{y}}^2 + 2g(h_0 - h_\mathrm{f})}}{g} \\[1.5em] x_\mathrm{f} = v_{0\mathrm{x}} t_\mathrm{f}

In the latter case, we have introduced the height of the landing ramp hfh_\mathrm{f} to have a more general result.

Car jump with air drag force

Let's introduce the air resistance into the computations of car jump trajectory. We'll use the most common form of the drag formula we described in the drag equation calculator:

Fd=bv2F_\mathrm{d} = - b v^2

where:

  • FdF_\mathrm{d} – drag force;
  • bb – air resistance coefficient; and
  • vv – car velocity.

The air resistance coefficient bb is equal to:

b=12ρACdb = \frac{1}{2} \rho A C_\mathrm{d}

where:

  • ρρ – air density;
  • AA – cross-sectional area of the car, seen from the movement direction; and
  • CdC_\mathrm{d} – drag coefficient that depends on the shape of the moving object.

The drag equation consists of several variables, which we often treat as constants for simplicity. However, it isn't true in general. The car rotates mid-air, meaning that the cross-sectional area and drag coefficient generally change their values during the jump. That drastically complicates the problem, and to find the exact solution, you might have to use numerical simulations we cannot provide.

The cross-sectional area of a car in x and y directions.

Still, in the car jump distance calculator, we assumed that b is constant, which is approximately the case if a car doesn't rotate very much. So, let's write the same differential equations as before but extended by air resistance contribution:

2yt2+g±bym(2yt2)2=02xt2+bxm(2xt2)2=0\frac{\partial ^2y}{\partial t^2} + g \pm \frac{b_\mathrm{y}}{m} \left( \frac{\partial ^2y}{\partial t^2} \right)^2 = 0 \\[1.5em] \frac{\partial ^2x}{\partial t^2} + \frac{b_\mathrm{x}}{m} \left( \frac{\partial ^2x}{\partial t^2} \right)^2 = 0

The ±\pm symbol means that there is ++ for the car rising and - for the car falling. The initial conditions remain the same as in the previous section, except for two additional conditions:

y(tmax)=hmaxytt=tmax=0y(t_\mathrm{max}) = h_\mathrm{max} \\[1.5em] \frac{\partial y}{\partial t}_{|t = t_\mathrm{max}} = 0