Probably the biggest schtick about the 2019 Mercedes G-Class is that it can drive right up a perfect 100 percent incline (lord knows how many times I saw it do its thing at the G-Class Experience at the Rolex 24)â€”a pretty tough feat of engineering that requires a lot of planning to perfect. But it is actually possible... given a relatively particular set of circumstances

Turns out, thereâ€™s a lot of pretty reasonable math you can do to figure out if a G-Wagon could climb that incline without actually risking it yourself, which Jason Fenske of Engineering Explained dives into here:

The first thing to note is that a 100 percent grade doesnâ€™t actually mean youâ€™re driving straight into the air at a 90-degree angle. Instead, itâ€™s a perfect 45-degreesâ€“or, the angle where the forward push of your acceleration and the drag of gravity back on your car equal out. If your car isnâ€™t balanced properly, youâ€™re either going to flip over or just fail to make it up any farther.

On the basic end of things, lots of technical components need to come together to make things happen: an engine and transmission powerful enough to handle driving that intense, a 4WD system to ensure an equal distribution of torque, tires that can actually withstand the forces at play and the way the weight is distributed throughout the car.

Fenske shows that itâ€™s a pretty finicky combination. If you donâ€™t have the right tires, for example (and he doesnâ€™t in the video), you throw off the calculations just enough that you might struggle to make the uphill crawl.

It gets a little more complicated when the whole â€śmathâ€ť thing comes in. It is, admittedly, not my strong suit. You can check out Fenskeâ€™s explanation of the calculations, but I asked our own resident engineer David Tracy to simplify it down for me, and this is what he told me:

Basically, if you have a car sitting on a slope, the force pulling it down the slope is the weight of the car times sin of the angle of the slope (itâ€™s the vector thatâ€™s pointing parallel to the slope). In this case, thatâ€™s one over the square root of two (.707), times the weight of the car.

Now, the force of friction between the tires and the road has to exceed that in order to prevent the car from rolling backwards. Friction is equal to the friction coefficient times the normal force (which is just the weight of the car times cosine of the angleâ€”itâ€™s the vector perpendicular to the slope). In this case, that means the force of friction is the friction coefficient times one over the square root of two (.707) times the weight.

So since 1/sqrt(2) times weight times the friction coefficient has to exceed 1/sqrt(2) times the weight, the friction coefficient has to be 1.0 or greater in order for the tires to provide enough grip to get the car up the slope.

This means the tires have to be able to accelerate the car at 1G. Why? Because a vehicle maximum acceleration occurs right at the limits of grip, and is thus set by the tireâ€™s friction coefficient. If we say the carâ€™s mass times acceleration is maximized when it is equal to the force of friction (which is the normal force (mass times gravity) times the friction coefficient which is greater than 1.0 in this case), then you end up finding that acceleration must be greater than gravity. Or greater than 1.0 G.

What Fenskeâ€™s doing here is testing whether he can get his accelerometer to show that, indeed, the car can pull 1.0 G. It appears to come close. He also mentions that, in order to avoid a tip condition, the force vector due to gravity (which acts through the center of gravity of the car) cannot be outside of the wheelbase, or youâ€™d end up with a â€śmomentâ€ť that would essentially torque the car about its rear contact patches.

His third point is that the vehicleâ€™s overall gearing and power means itâ€™s able to produce enough thrust at the wheels to overcome gravitational forces, and thus to climb the grade.