Dare Mighty Things. That was the hidden message in the parachute of the Mars Perseverance rover. It's not quite as mighty, but I am going to dare something myself: I'm going to try to figure out how far the descent stage would land from the rover.

OK, let me back up real quick. Just in case you don't know how this works, here is the basic landing sequence: The spacecraft entered the Martian atmosphere and then deployed a parachute. After that, a rocket-powered descent stage slowed the rover down as it approached the surface. At the very end of the descent stage, a cable lowered the rover to the ground. Then the descent stage used its remaining fuel to shoot away from the landing site.

It's this fly-away stage that I want to analyze. If I can get the acceleration as it leaves, then maybe I can model its trajectory to see where it would land. Yes, NASA knows exactly where it landed—they even have a picture of its crash site. But it's fun to see if I can do this just from the single rover video.

OK, let's get started. The plan is to use angular size of the descent stage to get the distance from the rover in each frame of the video. But what is angular size, and what does it have to do with position? Here is a quick experiment for you. Take your thumb and hold it at arm's length from your face and close one eye. Yes, really do this. Now find something in the room that your thumb covers up. What happens when you bring your thumb closer to your eye? It looks bigger and covers up even more stuff in the background. The actual size of your thumb didn't change, just its angular size.

Suppose there is some other object—maybe it's a stick of length L in your field of view. Imagine that you can can draw a line from your eye to each end of the stick. It would look like this.

Illustration: Rhett AllainThe stick is sort of like a part of a circle with a radius r centered on your eye. This means the length of the stick is approximately equal to the arc length that has an angle θ. Assuming the angle is measured in radians, then the following would be true.

Illustration: Rhett AllainIn case it's not clear, θ is the angular size of the object. If you know the angular size and the actual size (L), you can easily solve for the distance to the object (that would be r). Now, what if that stick isn't a stick but instead a Mars descent stage? See? This is going to work. I can just determine the angular size in each frame and use the size of the descent stage to get a value for the height of the vehicle.

The first thing I need to do is determine the angular field of view for the upward looking rover camera. I couldn't find the exact specs, so I'm just going to estimate it. Here is a frame with the rover hanging on the tether before landing.

Photograph: NASAAccording to NASA, the tether is 6.4 meters long—so I know the distance (r) in this image. Also, I can estimate the length of the descent stage (based on an image of it next to the rover) as having a width of 2.69 meters. With this, I can calculate the real angular size (as seen from the rover) with an angle of 0.42 radians. I can use that value to set the width of the entire video frame at an angular field of view (FOV) of 0.627 radians (this would be 35.9 degrees).

This is super useful. Now that I know the angular field of view, I can take any image and measure the angular size of the descent stage and calculate its distance from the rover. So I just need to find the angular position of the four sets of thrusters on the vehicle using video analysis software (Tracker Video Analysis). I did this for both pairs of thrusters to get the following position vs. time graph.

Illustration: Rhett AllainI'm actually surprised that this looks linear—but there you have it. My initial thought was that this would be a parabolic plot showing that this rocket stage was accelerating. It may indeed be accelerating, but with a very low acceleration, or it's possible that it has already fired its thrusters and is now just a free-falling projectile. But at least I can approximate the fly-away speed by fitting a linear function to the data and using the slope of the line. This works because velocity is defined as the rate of change of position, and this is a position-time plot. From this I get a fly away speed of around 8.2 m/s (18.3 mph).

But wait! There's more. It's clear that the descent stage is tilted at an angle. Of course this makes sense. The goal is for it to get a safe distance from the rover. If it just shot straight up, it would come back down and crash on top of Perseverance—that would be awkward. I can get an estimation of this launch angle. Basically, if I look at the apparent distance between thrusters in the direction of tilt compared to the actual distance I can calculate the tilt angle. Here, this diagram should help.

Illustration: Rhett AllainUsing the known distance from thrusters (front to back) and the apparent distance, I get a tilt angle of 52 degrees from vertical. I don't know if that's correct, but I'm going to use it anyway.

Mars Projectile Motion

Now we are ready for a real physics problem. It goes like this:

A Mars lander is performing a fly-away maneuver to obtain a safe distance from the Mars rover Perseverance. The descent stage fires its rockets to obtain a launch speed of 8.2 m/s with a launch angle of 52 degrees from vertical. If Mars has a gravitational field of 3.7 N/kg, how far from the rover will it crash? You can assume air resistance is negligible.

That's a great test question. Now for the answer. Yes, this is your basic projectile motion problem. The key is that the motion in the horizontal direction (I will call that the x direction) has a constant velocity, since there are no forces in the x direction. In the vertical direction (y direction), there is an acceleration of -g (where g = 3.7 N/kg) because of the downward gravitational force. Since the force is constant and only in the y direction, I can separate the problem into an x motion and a y motion. These two motions are independent except for the time it takes.

Let's start with the vertical motion. In the y direction, the descent stage starts with a component of the 8.2 m/s velocity (since it's moving in both the x and y directions). Here is a look at this vector velocity at the beginning of the motion.

Illustration: Rhett AllainOh! You thought the vertical component of velocity depended on the sine of the angle? Not in this case. Since the angle is measured from the vertical (instead of the horizontal), the vertical component is the adjacent side of the right triangle and you would use cosine. With that, we can use the following kinematic equation for motion with a constant acceleration:

Illustration: Rhett AllainBoth the initial and final y position are equal to zero (on the ground) such that we get the following expression for time:

Illustration: Rhett AllainNotice that if you start with y0 at about 6.4 meters (which is more realistic), then you would have to use the quadratic equation to solve for the time. It's not that difficult—you can do it as a homework question and see how it changes the final answer. But we can use this time in the horizontal motion of the descent lander.

Here is the equation of motion in the x direction.

Illustration: Rhett AllainNotice that the velocity depends on the sine of the angle, since it's the opposite side of that right triangle—right? Now I can just let x0 be zero and substitute my expression for time to get the following:

Illustration: Rhett AllainYes, there is a trig identity you could use here to simplify—but it's not critical. I have all the values, so let's plug in the numbers. With that, I get a distance of 17.6 meters. Alas, this is wrong. Using this annotated image from NASA, it looks like the descent stage landed about 1,000 meters from the rover. I wasn't even close. Obviously the descent lander was OK. It's cool, I'm just going to write a new physics test question. It goes like this:

The Mars decent stage for Perseverance needs to fly away from the landing to a safe distance of 1 km. The launch speed of the lander is 8.2 m/s with an angle of 52 degrees with respect to the vertical direction. How high should it fly vertically before shutting off its engines?

We can solve this one. I know it. Yes, I am making the assumption that the descent stage moves straight up before becoming a projectile (again, with negligible air resistance). In this case, I'm going to start with the x motion equation, since I know the final landing position (1,000 meters). From this, I can solve for the projectile time.

Illustration: Rhett AllainNow I can use this time in the vertical motion equation and solve for the initial y position (which won't be zero).

Illustration: Rhett AllainThat expression could be simplified, but I have all the values. I'll just go ahead and plug them in. This gives a vertical starting position of 43 kilometers. OK, this is also a silly answer—but it's still a nice physics question. Of course, the real answer is that the descent stage accelerated and increased its velocity while firing its rockets. This means that during that time it not only increased in speed but also moved down range. It's funny how you can start off with a problem that seems simple but actually isn't.

OK, last try. I'm just going to make a numerical calculation in Python. It's basically two stages. First, the rocket will fly with a constant acceleration at the angle of 52 degrees for some amount of time. Yes, I'm just going to pick the time and the acceleration. After that, it's just a plain projectile motion.

Here is the trajectory for a plot that seems to work. (It's actual Python code, so you can change the values if it makes you happy.)

Illustration: Rhett AllainFor this run, I have a rocket acceleration of 6 m/s2 with the thrusters firing for 7 seconds. The final position of the descent stage is 964 meters. Close enough. Finally.

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