# Is the forward acceleration of a parachute equal to its downward acceleration?

I'm writing a small simulation program about parachuting.

I know that the chute will accelerate to it's maximum fall speed at a rate of 9.81 meters per second per second. The forward velocity of the chute is dependent on the downward velocity. It will only move forward (separate from the wind) based on how much air is pushing upwards into the canopy, which is dependent on the downward velocity. The chute is designed to move forward when there is no external wind force, in the same way that a plane's wings can push it in different directions based on their angle.

Would it then be accurate to say that the forward acceleration is equal to 9.81 m/s^2, since it cannot move forward bereft of it's downward velocity?

Or, would I rather have to do some trigonometry to find the acceleration (like finding the bottom side of a right triangle using the other non-hypotenuse)?

• Ideally, an open parachute does not accelerate downward at $g$. :) In fact, once fully deployed, the parachute and its payload should reach a stable terminal velocity. Most modern parachutes are designed to be able to control flight somewhat, so there will be accelerations used to change direction and speed of descent, but, on the whole, forces should balance and acceleration should be much less than $g$. – Rick Goldstein May 13 at 18:01
• @RickGoldstein right, it's velocity is much less than normal (about 9 knots down), but why wouldn't it accelerate at the same rate? – Premier Bromanov May 13 at 18:28
• Acceleration implies a change in velocity. There is a gravitational force, $mg$, downwards, which would accelerate the parachute/payload downwards at $g$, but only in the absence of other forces--air resistance, in this case. Acceleration is due to the total force on an object, so you need to add up all the forces (as vectors--force has a magnitude and direction). – Rick Goldstein May 13 at 20:52
• @RickGoldstein That's essentially what Im asking. I have split the vector into 2 parts since I only care about down and forwards (x and y if that helps). Is the downward acceleration the same as the forward or not? – Premier Bromanov May 14 at 15:24

First, consider an idealized, cylindrically symmetric parachute falling vertically (think of an old style circular chute). The net force on the chute is $$F_{net}=F_{gravity} - F_{drag}(v)=mg -F_{drag}(v)$$ where the positive direction is taken to be downward, and $$v$$ is the instantaneous speed of falling. Since $$F_{drag}(v)$$ is generally a monotonically increasing function of $$v$$ (the faster you go, the more drag), there is some $$v$$, called the terminal velocity, $$v_T$$, for which $$mg=F_{drag}(v_T)$$, so $$F_{net}=0$$, and, by Newton's Second Law, there is no longer any acceleration downwards. Because there are no horizontal forces, there is no acceleration in the horizontal direction, and the parachute tends to fall straight down.

If we consider slightly different initial conditions, like initially jumping from a fast moving airplane, the initial horizontal motion relative to the air will cause an opposing horizontal component of the drag force. Since there are no other external horizontal forces, the drag will slow the parachute until its horizontal component of velocity relative to the air goes to zero.

Now, as the OP points out, modern parachutes are designed to act a little like gliders. The skydiver has controls that allow the shape to be adjusted in a way that redirects airflow around and through the parachute. So now we can model the net force as $$\vec{F}_{net}=\vec{F}_{gravity}+\vec{F}_{drag}(v)+\vec{F}_{flow}$$ where $$\vec{F}_{gravity} = mg\mathbf{\hat{y}},$$ $$\vec{F}_{drag}(v) = -F(v)\mathbf{\hat{v}},$$ where $$\mathbf{\hat{v}}=\vec{v}/v$$ is the unit vector in the direction of motion, and $$\vec{F}_{flow}=F_{thrust}\mathbf{\hat{x}}+F_{lift}\mathbf{\hat{y}}.$$ Here $$F_{thrust}$$ is the horizontal force caused by redirecting the flow of air from the front to the back of the parachute--by Newton's 3rd Law, redirecting the air horizontally means the air exerts an opposite force horizontally on the parachute. $$F_{lift}$$ is the vertical component of the force caused by the asymmetric air flow over the parachute, as in flow over a wing. These are both second order effects that depend on the speed of the system through the air, as well as details of the parachute shape.

When we add up all these forces, we expect the net force to be zero, since we ideally have a constant speed relative to the ground (both horizontally and vertically). We can separate the horizontal and vertical components of the force equation, but they will not be entirely independent. Redirecting airflow to increase horizontal speed will increase horizontal drag, increase vertical lift, and potentially reduce the vertical drag. In general, however, the biggest components of force will be gravity and drag, and the drag will have its largest component in the vertical direction. The horizontal force components will depend on the details of the parachute configuration, and can be anything within the performance limits of the chute.

So, the short answer is no. The horizontal force need not be the same size as the gravitational force (though lots of things about the system may ultimately depend on the size of $$g$$).

• Our chute is actually a circular chute – Premier Bromanov May 16 at 17:46
• @PremierBromanov I don't think that materially affects my answer. The ability to control the shape, however slightly, is what is required to adjust glide ratio and direction. If the system is cylindrically symmetric, we'd expect it to eventually reach an equilibrium of constant, purely vertical velocity with respect to the air. – Rick Goldstein May 16 at 19:15
• no it doesnt, i was just confirming that we do indeed use the circular one you suggested. – Premier Bromanov May 20 at 2:02