air resistance

The Curving Soccer Ball

rallain — Tue, 07 Sep 2010 08:11:59 +0000

The Curving Soccer Ball

You can call it football if it makes you happy. Anyway, this is a popular story going around. The physics of the magic curving soccer kick. Here are two ends of the spectrum.

First, there is the lower, easier to consume version from io9.com

Physics forced to come up with whole new equation to explain "impossible" soccer kick

I will summarize this article for you:

"Have you seen these crazy soccer kicks where the ball curves? It happens because the ball spins and physics. Here is a video"

Oh, and they have a diagram - which doesn't seem to come from the original paper and they also have some nifty real-life soccer videos. I think this story is a little too light on the details. They could have done just a little bit more to make this a much better article. Essentially they said that the ball curves because of magic (but magic is physics).

Then, there is the original article on the motion of spinning objects (which talks about soccer at the end) from the New Journal of Physics - IOP:

The spinning ball spiral - Guillaume Dupeux, Anne Le Goff, David QuÃ©rÃ© and Christophe Clanet

Let me select one tiny part of the paper to show you: (they used pictures for some of the variables, so some of this might not appear exactly as the author intended - but you will get the idea):

"The motion of the sphere of mass M is described in the Serret-Frenet coordinate system introduced in figure 2. We first focus on the direction . The Reynolds number Re = ÏU0 R/Î· is of the order of 104, which implies a drag F1/2ÏU2ÏR2Â·CD, with CD0.4 [28]. The equation of motion along thus is written as"

They lost me at "Serret-Frenet" coordinate system. So, this doesn't appear to be consumable for the more general audiences.

Update: While looking for some soccer ball data, I found a third article. The first was too cold, the second was too hot, but this one was just right for Goldilocks. This is from physicsworld.com.

The physics of football - Takeshi Asal

Like I said, I think this last article gives a better mix of understandablity and physics.

The missing parts

I am going to try to fill in the middle between the io9.com article and the original article. I might fail, but I am going to try. (even though the third article did a pretty good job)

So, you kick a ball. What forces act on the ball? Well, the easy thing is to say "gravity and stuff that touches the ball". In this case, the only thing the ball touches is the air. The air does indeed exert a force on the ball. The force the air exerts on the ball is ultimately due to collisions with the air particles and the ball. If the ball is spinning and non-smooth, there can be complex interactions. For this case, I will break this air force into two components.

Air drag. If you have read this blog, you should be familiar with this model of air drag that says the force is proportional to the magnitude of the velocity squared and some other stuff (density of air, cross sectional area, and shape of the object).
Magnus force. This is the force exerted on a moving and spinning object in a fluid or gas. Wikipedia's page on the magnus effect is pretty ok.

There is also the gravitational force. But, let me just look at the ball from the top view. The key point of all of this is that if there were no spin effect or air drag, the ball would just move in a nice parabola. From the top, this would look like a straight and constant speed trajectory. If you exert a force perpendicular to the direction of motion, the ball will turn. If you exert a force in the opposite direction of the motion, the ball will slow down. These two things together make the ball do what it does.

Here is a force diagram of the ball as seen from the top (so you don't see the gravitational force):

Why does this spinning cause a sideways force? Well, the idea is that the rough surface of the ball moves air near its surface. This means that on one side of the ball, the air is moving faster than the other side. On the faster moving air side, the air is moving more in a direction parallel to the motion of the ball. This means that an air particle is less likely to collide on the side of the ball and push it that way. The result is that there is more collisions on the slower side of the ball.

Modeling air interaction

Here is the model that is commonly used for the air drag force:

Where the v-hat is a unit vector in the direction of the velocity of the ball. This along with the negative sign means that the air drag force is in the opposite direction as the velocity.

The magnus force can be written as:

S is some constant for the air resistance of the ball (a basketball and a soccer ball would have different values). The vector ω is the vector representing the angular velocity of the ball. For the diagram shown above, the vector ω would be perpendicular to the plane of the computer screen and coming out of the computer screen. The mangus force is related to the cross product of ω and the velocity. (here are some cross product tips).

Why don't you always notice these forces? If the speed is slow and the mass is large, then the air drag and magnus forces will be small compared to the gravitational force. The motion for these cases will be dominated by the gravitational interaction. But with a high speed kick from a soccer ball (that has a relatively low mass) with a high angular spin, the effects can be noticed.

Let me model a high speed soccer ball in vpython. The original research paper gives some nice parameters that I will need for a soccer ball.

Radius = 0.105 meters
density = 74 times the density of air (if I understand the table correctly)
S = 0.21 - I am pretty sure the S in this paper is the same S in the magnus force described above. - forget this S

After playing around (and finding that third article) I am pretty sure the S above is not the same S as in the wikipedia page. The physicsworld article gives the following useful info:

Ball speed = 25-30 m/s
angular velocity = 8 - 10 rev/sec
Lift force (magnus force) of about 3.5 N
horizontal ball deviation of about 4 meters
ball mass of 410-450 grams (which agrees with my previous density)
ball acceleration of about 8 m/s² - not sure if this is just the linear acceleration or the total magnitude of the acceleration and at the beginning or average?

If I assume the magnus force is S times the cross product of the angular and linear velocity, I can work backwards to find S (from the physicsworld data) in the case that the velocity and angular velocity are perpendicular.

Now for some python (here is my sloppy code -

magnus_force.py). I will make one assumption - the angular velocity of the ball is constant (which obviously will not be true).

Here is what I get for the trajectory of the ball (as seen from above).

That is more than 4 meters deflection - but maybe they are assuming you aim to the left a little or something.

How about a plot of the total acceleration (magnitude) as a function of time.

This gives an acceleration of around 8 m/s² around the end of the motion. Maybe this is what the physicsworld author meant. Oh well, that is enough for this. I know there is one problem. I assumed a constant coefficient of drag, but it seems that this might not be true.

rallain Tue, 09/07/2010 - 04:11

Tags

air resistance

analysis

Physics

python

Serret-Frenet is a local comoving coordinate frame along a trajectory, particularly its vectorbasis description. It's not that crucial to understanding what's happening. Just useful for computations.

http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas

Very nice explanation. I've wondered about that before myself. I always wondered why the side of the ball that was spinning into the direction of velocity wouldn't feel more air resistance and cause the ball to curve in an opposite direction to what is experimentally observed. Thanks, well done!

Hi there, for my physics assignment i'm doing the physics of soccer and I would like to know the effect of spin on the ball when it is kicked

Hello, there are more examples of this "fenomenon" in soccer - in Poland there was a renown player, Kazimierz Deyna, who scored goals directly from a corner, from time to time. I believe on youtube one may find out some goals direct from a corner of other players.
Tom

watsssssssss uuuuuuuuuuuuuuuuppppppppppppp

What about the shape of the ball? Does that play into the direction at all, the ball must change shape a little bit when struck. Also, is this the reason when kicking a ball for distance, a little backspin will make it go further? If the air is moving faster on top of the ball in flight, it should stay in the air longer. And when you put topspin on a ball it dips.

Oh, and Puma came out with a ball with dimples, like a golfball. Do you know how effective that ball was, and what, if anything, was it good at? The dimples of a golf ball basically prevent drag, right?

Some more info on basketballs

rallain — Fri, 13 Aug 2010 10:10:57 +0000

Some more info on basketballs

You know I have trouble letting stuff go, right? I am still thinking about these crazy long basketball shots. Here are some more thoughts.

Really, there are two things I am interested in. First, commenter Scott Post suggests that the drag coefficient might be around 0.25 instead of 0.5. I don't know. For the discussion before, it doesn't really matter. My point was to see a numerical model for a falling ball would be similar to the time and distance from the video. Changing the drag coefficient to 0.25 gives values that are still close to the video. So, I still think the video is real.

The other thing that I would like to explore is the range of launch angles and velocities for a person throwing a ball. This would be important to have a feel for the spread of initial values so that I could estimate how many times you would have to try a shot to make it.

After a short search online, I found two interesting articles. One of them I know you can read - the other you might only be able to access if you are at an academic institution or if you are a subscriber to the American Journal of Physics. Below I will extract the useful details from these two papers.

The Perfect Shot - Seleh Satti

This pdf paper does not appear to be published in a journal, just online. The author looked at a few shots of a basketball indoors from the same distance away from the goal (5 shots). It seems the main point was to determine if you would need to include air resistance and spin effects in the analysis of this shot. To do this, Saleh used a video camera to capture the 5 shots (one of which was a miss) and then did some video analysis.

For air resistance and spin, the author claims there are not noticeable effects - mostly because there is no horizontal acceleration. It is odd that the vertical values for acceleration were around 9.1 m/s². But maybe this was just a scaling problem.

Here is a table showing the initial velocities for the 5 shots. This can give some idea of how consistent a shooter could be (although it would be nice to have more than 5 data runs).

That could be useful.

Physics of basketball - Peter J. Branczio

This paper is from the American Journal of Physics (didn't I already say that?) and really tries to do a lot of stuff. Part of it covers bouncing of a rotating ball (which is interesting, but not what I am looking for). The author also discusses the kinematics of projectile motion (but not with air resistance). There are some more details about projectile motion for an object of finite size that has to hit a target. This, in particular, diagram is interesting:

Basically, the lower the angle of incidence for the basketball, the smaller the apparent size of the hole. And here is another great diagram showing the range of initial velocities and angles that will result in a successful shot (where it doesn't hit the rim).

Finally, the author compares no-air resistance with air resistance for some cases. The result is that you would have to adjust the initial velocity by about 5% to compensate for the drag.

Where does this leave me?

The problem with both of these papers is that they are for normal basketball shots - not super duper shots, so air resistance isn't such a big deal. I would really like to have an experimental value for the drag coefficient of a basketball. Probably the best bet would be to get a nice video of a basketball traveling very fast. From this, I should be able to get the coefficient (I will put this on my to-do list).

The second paper was very detailed, but it approached the problem from the wrong direction. It said "what initial conditions would you need to hit the goal". I want to know what variation a player would have in throwing a ball. Oh well, I think I can get this data myself also.

rallain Fri, 08/13/2010 - 06:10

Tags

What about the backboard? I'm not much of a basketball player, but I do recall that it's supposed to be easier to make the shot if you bounce it off the backboard.

Well, I found that a different basketball shot that The Legendary Shots took was digitally manipulated, so it makes all their shots questionable. Take a look at http://jorfer88.blogspot.com/2010/08/irrefutable-evidence-that-legendar… .

Basketball shot - real or fake?

rallain — Sat, 07 Aug 2010 10:11:00 +0000

Basketball shot - real or fake?

I have seen several videos similar to this.

Real? Fake? How many tries did this take? Let the analysis begin. Before I do any analysis, let me state that I think this is not fake. I do not know that for sure, just my first guess.

How would I tell if it is real or fake? This is tricky. I can't really get a good trajectory of the ball to make some measurements on it because of the camera angle (next time people, make sure you set the camera up perpendicular to the plane of motion and far enough away to avoid perspective problems - thanks!) Really, the best I can do is to look at the time. How long does it take for the basketball to get to the goal? How hard would it have to be thrown to do this?

Gathering Data

Luckily, I know where this takes place - Vulcan Park in Birmingham Alabama. According to the Vulcan website, the statue is 56 feet tall and sits on a 124 ft pedestal.

So, how high is the ball thrown from? Here is a shot of the statue plus its base.

Using this and Tracker Video (yes, you can use it to analyze images too), I get that the walk-way is at about 95 feet above the ground. If the basketball is 10 feet from the ground (as normal), then the vertical height for the ball to drop will be about 85 feet (26 meters).

How far is the goal horizontally? This one is a little tougher. First, let me start with this frame from the video.

Pay attention to the light post and that tree. Using those two, I think I have pinpointed the spot of the goal.

Here is another shot from another angle.

Using this location, Bing Maps and Tracker Video the goal was probably about 130 feet (40 meters) horizontally from the throwing place.

I need one other thing. Information about the basketball. According to Wikipedia, the NBA basketball has a circumference of 29.5 inches and weighs 22 ounces. (this is a radius of 0.12 m and a mass of 0.62 kg)

Time Measurement

How long was the ball in the air? Again, I can use Tracker Video - even if I don't know position data, I can still get time. From the time it left the throwers hand until the time it hit the goal was around 3.43 seconds. For an added bonus, I have a couple of frames where the guy was throwing the ball. Plotting the position for this time I get (I scaled it with the diameter of the ball being 24 cm):

This is a plot of the total magnitude of the velocity as a function of time while the ball is being thrown. From this, the ball was thrown with an initial speed of around 10 m/s (22 mph). That seems reasonable.

Do I need air resistance?

I suspect I do, but let me do a quick calculation. Suppose there were no air resistance. In this case, how fast would the ball be moving at the end of the throw? I will use work-energy and take the system as the ball plus Earth (so there will be gravitational potential energy). This gives:

If the ball is moving at 24 m/s, how much air resistance would there be? How would this compare to the gravitational force? I will use the following model for air resistance.

Where ρ is the density of air, A is the cross sectional area of the ball and C is the drag coefficient (about 0.5 for a sphere). So, at this speed the air resistance and the weight are:

Since the air resistance force for this mythical final speed is greater than the weight, air resistance is not negligible. This means the easiest thing to do is a numerical calculation (intro to numerical calculations) with python (here are some numerical calculation examples).

For my numerical calculation, I will have the ball start 26 meters above the target and moving at 10 m/s. Looking at the video clip, I am going to use an initial launch angle of 21 degrees. Let the calculation begin.

Numerical Calculation

Here is what I got from vpython.

I added the ground and the tower for effect. Two important things from this simulation.

Time: The time for this ball to make its motion is 3.01 seconds. That is good. It is not exactly the same as the value from the video, but my parameters could be off a little bit. This is close enough for me (for now).
Distance: I looked at the map and the video again. I am fairly certain about where the goal post is. It should be about 40 meters from the starting location. However, when I ran the simulation, the ball only ended up 20 meters away. This is not good.

Ok, there is a problem with the distance, but maybe my starting parameters are wrong. What if the ball were thrown at 20 m/s? How far would it go? This is a simple thing to change. If I do that, the ball is in the air for 3.4 seconds and goes 37 meters. Could the ball be thrown that fast? Maybe. I admit that my initial value for the velocity was just based off of a few data points from tracker video. Also, there was a scaling problem and such. I don't put too much weight on that speed.

There is some other data I can look at. Using the camera from the shooter location, I can see the ball as it enters the goal. This is far enough away and the ball is mostly moving downward that I can get an estimate of its final speed and compare that to the numerical calculation. Here is a plot of the y-position of the ball at the end.

This shows the ball moving at a fairly constant downward speed of 17 m/s. In the numerical calculation, at the end of the run the ball was moving horizontally at 5.5 m/s and downward at 17 m/s. That is pretty close.

Conclusion

I am going to say not fake - although I have been wrong before (remember the giant water slide jump?). My reasons are:

I can model this motion and get a similar time (with air resistance)
The ball should be in the approximate area with an initial speed somewhere around 20 m/s
The final speed of the ball in the video and in the simulation mostly agree.

I am stopping here - but not forever. There is another very interesting question: how hard is this to do? Did these guys have to try like one million times or did they just get lucky (or are they that good?) I am saving this for another post.

rallain Sat, 08/07/2010 - 06:11

Tags

Nice post. But I think you meant to leave off the last line, seems like a note to yourself while you were writing.

@The Virtuosi,

HA! You are right. It was a note to myself. I removed it. Thanks!

Tee hee, you don't have to fake it. Just show the 1 out of 1000 takes that made it.
Is kinda like evolution, toss out what doesn't work, keep what does.

I suspect fake. I think, although this is just guessing, that the impact on the goal would o be a lot more violent than shown from that height. If you missed and hit the ring it would almost certainly break, and considering the number of tries you'd need to get that one, perfect hit, it would be an expensive video.

thanks, I wonder what the margins of error are in the angle thrown or initial force.

question:

if the ball landed on a young squirrel would it kill it?

the ball hits the net 3 meters above the ground.

I am not sure how much difference it would make, but I have read that the drag coefficient of a basketball is actually 0.25. Like a golf ball, the dimples trip the transition to turbulence, reducing the wake effect. I have not seen experimental proof of this value of drag coefficient. It is on my list of things to test.

It really doesn't look like he threw it that hard, but athletes can put a lot on a ball with little effort. I've seen a college bench player make a turnaround jumper from the sideline at half court.

I would position the goal after throwing a few basketballs off of the tower. Then you are relying on the ability of an amateur athlete to repeat a physical action and only have to worry about windage.

The ball is traveling close to vertical, which helps with the margin of error. The jumpers LeBron can make from the baseline (about 94 feet) are tougher.

Impact of the ball on the rim is nothing compared to a dunk. These portable baskets are pretty tough.

I'd suggest some model confirmation experiments, both real (drop a ball off of a tall building or take a road trip to that tower) and video (tracker analysis of LeBron's baseline shots). You also have the apex of the ball's flight (the form line on the elevator tower above the door) and the time to that point to calibrate your initial speed/angle estimate.

@Scott,

Good point. Really, someone needs to measure the coefficient for a basketball - or maybe this has already been done? If the coefficient is much smaller, he wouldn't have the throw the ball as hard.

You mentioned your values for initial speed, but you didn't give us the throw angle. Did you just simulate a horizontally thrown ball? I'd estimate the angle at 10 degrees or more above horizontal.

Adjusting speed and angle separately could allow you to keep your fall time accurate while getting your horizontal distance closer to the theoretical value.

(I should say that other commenters have mentioned the throw angle, but I want to address that explicitly.)

Looking at the video clip, I am going to use an initial launch angle of 21 degrees.

D'oh. Of course as soon as I post my comment I find where you answered my question. Reading skills. I need them.

@9:
You can learn all you need to know by dropping it about 100 feet and analyzing the video to find the velocity dependence of the acceleration.

I know that genuinely skilled people can put a lot more into a throw than a neophyte, but 20 m/s is 45 mph. As I recall from an experience with a pitch radar at a ballpark one time, I can throw a baseball in the vicinity of that speed by putting my entire body into it. (I know that professionals can throw twice that fast, but they're one-in-a-million freaks of nature (in the good way) who still, eventually, pull their own joints apart as often as not.) I'd characterize the video dude's throw of that basketball, which is much heaver and more awkard than a baseball, as a "casual toss."

Let's try a more quantitative approach. I've seen (real) videos of people sinking full-court shots. They look like they're throwing the ball with a whole lot more effort than what video dude was putting out. A basketball court is aobut 55 m long. Assuming a player throws the ball up at a 45-degree angle (probably not optimal for making the basket, but let's try it as a first guess), and neglecting air resistance (again, might be a problem), the player would need to throw the ball with a velocity of sqrt(((9.8 m/s^2)*(55 m))/4) = 11.6 m/s. At a (more realistic?) steeper trajectory of 60 degrees above horizontal, then the required velocity is sqrt(((9.8 m/s^2)*(55 m))/(sqrt(3)/2)) = 24.9 m/s. This is within the ranges of speeds shown in the video, but again, good basketball players look like they have to work pretty hard to put the ball up with these kinds of velocities, and video dude is not working that hard.

Barring some revision to Rhett's modelling, I'm going to guess that the video is fake.

OK, for all you skeptics out there, first required viewing before you debate me here> http://www.youtube.com/watch?v=lhbISRqSXW0. I pose this. Why would these dudes go through all that, then fake the swish shot? They shot from 2 angles and you can watch the ball leave the tower and connect with the net. I'm just a layperson and simple math tells me at what speed an object falls to earth and the time it takes to get there. The seconds from when the ball leaves the tower to the net seem correct.If you still think that the dudes faked the shot, then email or call fox6 news and have them recreate the shot and let fox6 film it.

Rhett: to alleviate the non-perpendicular to motion camera angle, can't you do some co-ordinate transformation to estimate what the projection of the true trajectory would look like from the camera angle?

I think that mapping the position of the shadow of the ball in the ground may give some extra information about the trajectory to compare with the model, as a double check.

I'm the sports anchor at the Fox TV station in Birmingham. The Vulcan statue is right next door to the station. We had two news photographers video tape these kids- it took them about 200 tries but yes, they made the shot, and no,the video is not doctored.

@rob

I have tried something like this before with limited success. It is a pain.

@Esteban,

Wow - I didn't even think about the shadow. Great idea. I am not even sure how well it can be seen in the video, I will have to go back and take a look.

@Rick,

200 tries? That is a very important piece of data - it will be useful later. Thanks!

Did they get closer as the shots progressed?

The elevation from the base of the monument to the elevation of the goal was an additional 10 feet. I am the mom of the kid who is the founder of the Legendary Shots which made this Vulcan shot. Can you compare this the the Dude Perfect shot at the Texas A&M football stadium?

@Jill,

I have seen the Texas A&M shot, and I wanted to analyze it. However, I found it was easier to get specifications of the distances with the Vulcan statue than the Texas A&M stadium. If someone has a few more details about that shot, I can probably put it together.

all i have to say is that there is no way of that ball having a 20m/s - 72km/h initial speed but i think that it isnt fake.

Speed of a rising drop of oil

rallain — Fri, 25 Jun 2010 09:16:14 +0000

Speed of a rising drop of oil

The oil spill is still in the news (sadly). One thing that keeps coming up is the speed that the oil bubbles rise to the surface. This is important in different oil-capture methods. The common statement is that smaller bubbles of oil can take quite a long time to reach the surface and larger bubbles can take about 2 days.

This is one of those cases where things do not scale quite the same. Suppose there is a spherical oil bubble rising at a constant speed. Here is a force diagram for such a bubble:

If this drop is going at a constant speed, then all these forces have to add up to the zero vector. That is fine, but here is the interesting part. Let me describe these three forces:

Gravitational Force

Close to the surface of the Earth, I can just say this force has a magnitude of mg where m is the mass of the drop and g is the gravitational field (9.8 N/kg). The mass is the interesting part. If I assume an oil density of ρ_oil and a radius of r, then the mass would be:

The main point here is that the weight is proportional to r³.

Buoyancy Force

I am not going to go into the details of the buoyancy force (but here are some posts on that topic). Let me just say that the buoyancy force depends on the volume of the oil. So it also has a dependency on r³.

Drag Force

Is this drag force proportional to the velocity or the velocity squared? You know what? It doesn't matter. What matters is that it depends on the cross sectional area of the oil drop. The bigger the drop, the bigger the drag force. Suppose that this drag force is proportional to the velocity, then I can write the magnitude as:

Maybe you can already see the point. This force depends on the radius squared. If I put all of these forces together and solve for the velocity, I get (these are just the y-components of the forces):

There you have it. Since the buoyancy and the weight essentially depend on the volume (r³), but the drag depends on the area (r²) the r-dependency doesn't go away. Instead you have a terminal speed that depends on the size of the drop.

Our common intuition says that if you make a bigger drop, all things should be bigger to make the same effect. However, this doesn't always work. If you double the radius, the volume is 8 times bigger, but the cross sectional area is only 4 times bigger.

rallain Fri, 06/25/2010 - 05:16

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Star Trek Space Jump

rallain — Mon, 24 May 2010 12:54:46 +0000

Star Trek Space Jump

While I am still fresh on the Space Jump topic, let me take it to the extreme. Star Trek extreme.

SPOILER ALERT

But really, is this a spoiler alert if it is from the trailer of a movie that has been out forever? Of course, I talking about the latest Star Trek movie where three guys jump out of a shuttle and into the atmosphere.

So, in light of the Red Bull Stratos jump, how would this jump compare? First, my assumptions:

This Star Trek jump is on the planet Vulcan. I am going to assume this is just like Earth in terms of gravity and density of air.
The jumpers in Star Trek have on stuff that is different than what Felix will wear in the Stratos jump - but I will assume these guys will have similar falling characteristics.
The jumpers start from a low orbit similar to the orbit of the space station. I will use a starting height of 300 km above the surface.
The jumpers are not in orbit. I will assume their initial starting speed is 0 m/s.
The model I am using for the density of air is only valid to about 36 km above the surface of the Earth. Higher than that, I am just going to have to estimate the density of air (see below)
The drag coefficient is constant. This is really not true, but it is the best I can do. Sorry, I will try harder next time.

Ok, now what do I want to look at? I will compare this Star Trek jump to the Red Bull Stratos Jump in several ways:

Maximum acceleration
Maximum speed
Speed compared to the speed of sound

Density of air

Since my model for the density of air seems to only be valid up to 36 km, I need to do some thing else for the other 250 km. My first thought was jut to put the density at zero. But then I thought that might not be the best thing. Even a very low density can make a big difference dropping that first 250 km. Here is a graph from Wikipedia showing the density as a function of height.

Actually, I have a new plan. This was not trivial to find (lots of broken links) but here is NASA's MSIS-E-90 Atmosphere Model. What a find. Using this I can generate air density as a function of altitude to 300 km. Here is a plot of that data:

And here is a plot of the old density model I used in the last Red Bull post along with the new NASA approved one.

Those are close enough for me. I will just use the NASA-Navy model (well, I will use select points from that model).

Maximum acceleration

I already did this for Felix and the stratos jump. Here is what I got:

So, not too bad. The maximum acceleration is less than 1 g. He could easily handle that (even I could). Now, for the Star Trek guys, I just need to change the initial height to 300 (and change the density model).

This looks crazy. Part of the problem is that in order to get density data over 300 km, I had it broken into big chunks (10 km sized chunks). Obviously, that is too big. Also, another problem. The acceleration never goes to zero. This means that the jumper wouldn't reach terminal velocity. I just don't think that would happen. Even meteors usually hit terminal velocity (I think). Here is what I am going to do. I am going to use these big chunks for stuff greater than 39 km and then use the old Red Bull way of calculating the density for stuff below that. Doing that, I get:

I like this one better. There might still be a problem with the density around 39 km. I am a little worried about the sharp increase in acceleration. I changed my density model so it was much more "detailed" at the higher altitudes. I am still using the old density model for heights less than 30 km.

So, what does this mean? This means that for most of the jump (above 39 km) there is so little air resistance, the jumpers just super speed up. Like ZOOM. After 39 km altitude, the air resistance really starts to increase. It is almost like hitting a wall since they are falling so much faster than terminal velocity. This make the air resistance force ginormous and the resulting acceleration deadly. Well, maybe not deadly. The Wikipedia g-force tolerance page says that an acceleration of 25 g's is possible for about 1 second. However, in this fall, the jumpers will have over 20 g's for over 4 seconds. Maybe they have special Star Fleet issue suits that allow them to experience higher accelerations. I mean, if they can make inertial dampeners for a ship, they surely can do this.

Maximum speed

Now that my air density model seems to be working well enough, it is relatively simple to look at the speed of the star trek jumpers.

Top speed just over 2,200 m/s (4900 mph). In physics, we call that zoom-fast. Remember that from 120,000 feet, a jumper would get around 250 m/s.

Comparing the speed to the speed of sound

If I use the most basic model of the speed of sound, it only depends on the temperature of the gas. This is a problem when you get up to 300 km above the Earth. So, instead of plotting the speed of sound, I am just going to calculate the speed of sound at the height where the jumper will be going the fastest. From the previous plot, I get a max speed of about 2,200 m/s at about 36,000 km. The speed of sound at this height is about 200 m/s. The answer to the question: the star trek jumpers are going way faster than the speed of sound, about mach 11.

Ok - I think what I need to do is to implement NASA's atmospheric density model in python rather than discreetly take data points from their online thing.

rallain Mon, 05/24/2010 - 08:54

Tags

red bull stratos jump

Problem with your assumptions: It is a matter of canon that Vulcan has a significantly higher gravity than Earth.

Of course, given the utterly ridiculous physics in most of that movie, it's sort of amazing that the jump scene wasn't just plain wrong on every possible level.

I mean, if they can make inertial dampeners for a ship

A sponge is a dampener. You're probably thinking of something else.

(That's one for the pet peeve file.)

The real problem with the high accelerations is that they are going head first. No if they were going feet first there would be no problem. For very high accelerations they have seriously strong arch supports.

@Nick,

Oh - is that why Spock is so strong? Because of the higher gravitational forces?

And yes, I was surprised as well that it wasn't so far fetched.

@Keith,

No - according to wikipedia's g force tolerance entry humans are best when the acceleration is perpendicular to the spine - "eye balls in".

Hi Rhett,

Is there a problem with labels on the third graph? It says "hight [km]" which would make 10 thousands, 20 thousands and so on km ...

Also on Maximum Acceleration you say "The maximum acceleration is less than 1 g", when I think you mean 10 g.

Sorry for the nitpicking, but I try to follow the numbers in your posts and it gets confusing.

@Jon,

Thanks for pointing out the mistake on the graph. I fixed it.

For the "less than 1 g", the above graph is the acceleration in m/s^2. His max acceleration is less than 10 m/s^2 which is less than 1 g (where 1 g = 9.8 m/s^2)

@Rhett,

Yes, the high gravity of their home planet is the canoncial reason for the great strength of Vulcans. I'm not entirely sure that makes sense, but there it is. (Wouldn't they also be shorter and more heavily-built if this were so?)

According to the Memory Beta wiki, surface gravity on Vulcan is ~1.4g. There are also references to a thinner atmosphere, as well. (Maybe that's how they were able to hang a thousands-of-meters long cable in it without any detectable sway or buffeting? Oh, wait, they did that over San Francisco as well. Never mind. Seriously, the science in that movie was so excruciatingly terrible that it made me long for the "quantum positronium tachyon soliton flux" nonsense of ST:NG.)

"Wouldn't they also be shorter and more heavily-built if this were so?"

Generally assumed that would be the case.
But do we know whether Vulcans evolved there? Maybe they already are shorter and stockier than their ancestors who might have arrived on Vulcan from some other planet with much lower gravity.
Loss of historical data about that could well happen (or maybe they were placed there as lifestock or slave pens in pre-technological era, anything is possible).

Back on topic though. Do we know whether those jumpsuits don't contain similar inertial damper technology as is contained in federation spaceships? This is a movie after all, and one that doesn't take physics too seriously.
One can envision a system where inertia is somehow converted into heat and radiated out through some mechanism unknown to us.

That would reduce the decelleration forces experienced by the jumper to survivable numbers without leaving the technological framework establised by the movie scenario.
After all, we're talking Hollywood here, not Newton.

Stratos Jump for Mere Mortals

rallain — Mon, 24 May 2010 08:35:00 +0000

Stratos Jump for Mere Mortals

Crazy, but I was on CNN Saturday night. They contacted me at the last minute to talk about the Red Bull Stratos Jump. Here is a screen shot to show that I am not making this up (or that I have awesome photoshop skillz).

Looking back, maybe I looked like an idiot. Really though, it wasn't my fault. I thought we were going to talk about physics. The first two questions threw me for a loop. Here are the two questions and my response (roughly paraphrased):

Will Felix survive the jump?

Answer: I guess so.

Is there a scientific reason for this jump?

Answer: I thought we were going to talk about physics. So....maybe?

Maybe it wasn't actually that bad. However, thinking about this, I want to give it another shot. So, what are the key take-home points I would like the general public to know about the Red Bull Stratos Jump? (in no particular order)

Forces and terminal velocity

There are really only two forces acting on the skydiver as he or she falls. There is the gravitational force - in this case it is essentially (but not exactly) constant. Then there is the air resistance force. This is a force from the skydiver colliding with the air. A couple of key things about the air resistance force:

It depends on the density of air. This is important in this case since the density of air changes with altitude.
It also depends on the surface area of the object as well as the shape. How about we just assume these don't change.
It depends on the square of the velocity in the air.
The air resistance force is always in the opposite direction as the motion in the air (so, for this case, that will always be up).

There are really only three ways these forces can combine resulting in three different types of motions.

The key thing about forces is that they CHANGE the velocity of an object. If the total force is zero (like case C) then the velocity will not change. For a skydiver, this is called terminal velocity. Normally, a skydiver starts the jump at around 10,000 feet. Sure the air is thinner up there than at the ground, but not THAT much thinner. This means that the skydiver quickly reaches a point where the air resistance is equal (but in the opposite direction) as the gravitational force and travels at a constant velocity the rest of the fall.

The key difference for the Stratos Jump is that Felix will start at an altitude where the density of air is really really small making a small air resistance force. This also makes for a very large terminal velocity (you would have to go super fast for the air resistance to be as large as gravity). During the time period A the total force is in the same direction as the way the jumper is moving, so that makes the jumper speed up.

As the jumper gets into higher density air, the air resistance force gets really large really quick. This makes the air resistance force much larger than the weight. Now (during time period B above) the net force is in the opposite direction as the motion of the skydiver. For this case, the skydiver will be slowing down.

Eventually, the speed will slow down making the air resistance force smaller so that they are equal (time period C). If there are equal forces in opposite directions on the skydiver, this is the same as no forces. If there are no forces on an object (or no net force) the velocity will be constant. This is terminal velocity.

How fast will he go?

The common answer to this question is that Felix will not go that fast because the fastest a skydiver can fall is around 200 mph. This true for normal skydivers where they change their body position to have a smaller cross sectional area. This means that for the air resistance to be equal to the gravitational force, they have to go faster.

For Felix, jumping from 120,000 feet, this doesn't hold true. The key difference is the very low density of air that will allow him to go super fast. He could reach speeds near 700 mph.

Faster than sound

This is a tricky question. The key thing here is "what is the speed of sound?" In the most basic model of gases, the speed of sound only depends on the temperature. So, as you go higher and the temperature goes down, so does the speed of sound. Felix will not have to go 740 mph (the speed of sound at sea level) to break the sound barrier.

Will the forces be too great?

For this particular jump, there will be a larger than normal air resistance force (see above) as the skydiver transitions from going very fast in low density to higher density air. If you start at 120,000 feet, this air resistance force will produce an acceleration less than 2 times normal gravitational feelings (2 g's).

What about the science?

This is the question I failed at in the interview. But I am ready now. Is there any scientific reason to do this? The best answer may be that science can be found everywhere. Think of all the things man has done that produced some cool scientific idea. These are not always planned experiments. The key is to just keep your eyes open and observe. You will never know what you will find.

From an engineering view, this jump will test some useful stuff. How do you get a man so high in the atmosphere? How about the spacesuit? How about the parachute? Also, maybe he can collect some atmospheric data. What about human performance at such a high altitude?

Finally, from a learning viewpoint, I think this is a great problem for introductory physics courses. Oh - Red Bull, please collect and share acceleration data during the fall. Please?

More details

Is this not enough? Do you want more details (in terms of the physics?) Here are a few posts you might like:

rallain Mon, 05/24/2010 - 04:35

Tags

I'm on to your games now Rhett! From Makezine. If they had been true blog readers though they would have interviewed you after the jump and then had you use Tracker Video to break things down.

@Jason,

Wow - that link is awesome. I am going to have to try that.

Next up - Rhett on the Today Show!

Based on absolutely nothing but WAG, I'd guess that terminal velocity for a human body (padded by altitude suit) would be well short of Mach 1.0 -- and the air temperature would make that pretty slow.

There's still enough air at that altitude to keep a subsonic aircraft (think U-2, although they don't go quite that high) up, for instance. The density is down to the point where rolloff is pretty slow, too.

They contacted you and you assumed it was as a physicist. Non-physics questions gave you a little pause. In future you might do better to realize that their audience is a bunch of beer drinking dopes whose knowledge of physics is limited to some vague recollections of people dropping cannonballs off of towers and that physics had something to do with making the bomb.

In other words you, being a physicists, has little to do with it. Your a stand in as the all-purpose 'science guy'. And talking to an audience of poorly motivated folks operating at middle school level.

You really did a good job of it. But you would have done a better job if you had chugged a couple of beers beforehand to take the hard science edge off and tune you in to the beer addled, couch potato frequency that TV operates on. I say chugged because a loud belch early in the interview would endear you to the TV audience. As long as you make only a cursory effort to apologize and carry on as if nothing had happened.

You have to realize that the audience is amotivational 16 to 25 year-old and the main uses for Red Bull is to keep the drunks awake so they can drink more, and to get hungover people moving again.

Transcript. Hey-o!

-------------------

LEMON: Why? I have to ask him that. I wish I could ask him that question. You know, this seems like the craziest science experiment ever, right? So we asked a scientist to help us understand what Baumgartner is trying to do. So Rhett Allain is an associate professor of physics at Southeastern Louisiana University. And he joins us from Hammond, Louisiana, tonight.

Good to see you, sir.

Is Felix, Fearless Felix going to survive this?

PROF. RHETT ALLAIN, SOUTHEASTERN LOUISIANA UNIVERSITY: He probably can. I mean, Joe did it from 102,000 feet. 120,000 feet will be a lot faster, but he'll experience a lot of the same things. There won't be any forces outside of the realm of what a normal person should be able to take.

LEMON: I would imagine there is some -- a scientific reason to do this. Maybe it helps us understand you know what it's like to do something. Maybe it helps NASA. I don't know. I don't understand why someone would want to do this.

Is there a scientific reason to do this?

ALLAIN: Oh, I don't know about the science of this. I mean, I approach this from the science of how can we apply basic physics that a college freshman would look at in terms of this case. Atmospheric science. Maybe there is something there. That is not really something that I'm too much of an expert in.

LEMON: Yes. What I'm trying to get to is there a scientific reason to do it or is this just for the rush or the adrenaline to see what it feels like. By the way, he is being sponsored by Red Bull. We should say that.

So what is it like to fall so far, so fast?

ALLAIN: Well, I mean, the thing that is different here is at that high altitude, the density of air is very low. So you have these two competing forces, gravity and air resistance. And with the very, very low density, the air resistance isn't that much so you can really go real, real fast until you get to lower atmosphere and then you start slowing down. So he can go significantly fast.

LEMON: All right.

Hey, listen, it is good to see you. Thank you so much. Appreciate you spending a Saturday night with us. And we wish Felix, Fearless Felix Baumgartner the best of luck. God speed as I say. We hope he is safe.

Thank you very much, sir.

ALLAIN: Thank you.

-------------------

@Dan,

Where did you find that? Or can you just type really really fast?

Dan's rendition of the interview is definitely a bit off in some important details but maybe I'm just a perfectionist. Anyway, I like Art's post. I'd like to add that I was watching the interview with my sister (not a physicist btw) and she initially replied that you didn't answer the question after the interviewer's 2nd question. I immediately paused the video and replied, in your defense: "Did you hear that question (or series of questions)? Did you hear how broad and vague it was? I wouldn't know how to begin to answer it either?" Apparently all my sister heard was: "Blah blah.... question about scientific reasoning... blah blah.... How will it feel for Felix to fall from such a height?" This is the question (based on my sister's non-scientist point of view) that American's heard when they watched this interview. So I guess Art is right. A good way to interest the average unmotivated American would be to start with saying a joke like "It would feel awesome!" Hind sight is 20:20, I know. But then you could go into the details of forces acting on Felix and show how much academic value you have in this context.

@Patrick,

I think Dan's rendition is the official CNN transcript - http://www.cnnstudentnews.cnn.com/TRANSCRIPTS/1005/22/cnr.09.html.

And yes, this could have been so much more - a great learning opportunity. But instead it was a great PR opportunity, and I thank CNN for that.

Thank you for the tip. Looks like I need to listen more when these guys are trying to teach me skydive speak...

How hot would the space jumper get?

rallain — Wed, 19 May 2010 12:56:30 +0000

How hot would the space jumper get?

A new video from the Red Bull Stratos Jump guys came out. Here it is:

This reminds me of an unanswered question about the Stratos jump that I didn't address on my last post on this topic. Commenter Long Drop asked about how much Felix would heat up as he falls from 120,000 feet. This is a great question. The first, off the bat answer is that he won't heat up too much. Why do I say this? Well, when Joe Kittinger jumped from over 100,000 feet and didn't melt. Still, this is a great thing to calculate.

How do you calculate something like this? I will look at this in terms of energy. For simplicity, I will consider the jump from 120,000 feet down to about 30,000 feet. After that, Felix will pretty much be a normal sky diver. Here is the plot of speed vs. height from my previous post.

Just a note, the green line is the speed of sound, the red line is his speed if he jumped from 100,000 feet and the blue is from 120,000 feet. Think about this fall. If there were no air resistance, he would be going much faster and would have much more kinetic energy. So, without air resistance I could use the work-energy principle. If the Earth and the jumper are in the system, then:

But with air resistance, the jumper will not actually be going that fast with that much kinetic energy. So the missing energy had to go into an increase in thermal energy. This increase in thermal goes both into heating up the air and the jumper. But, how much goes into the air and how much into the jumper? I am just going to make a basic assumption that half of the energy goes into the air and half into the jumper. Simple, right? Now I just need to re run my numerical calculation and get the difference between the no air kinetic energy and with air kinetic energy. Here is a plot of kinetic energy vs. height (both with and without air resistance).

From this, I get the values:

That seems like a lot, even if only half of that went to the jumper. Instead of calculating the change in temperature, let me think about this in terms of power. That can give me the change in thermal energy, but how long did it take? From the numerical calculation, falling to 30,000 feet takes about 150 seconds. This would give an average power (so I could compare to an electric heater) of:

Still not very good. I just can't imagine having a 70,000 watt heater hooked up to you for even 2 minutes. Maybe the time is so short, it doesn't matter. Here is an idea. What if I do the same thing for a normal sky diver? Let me assume a sky diver jumps from 10,000 feet to 3,000 feet falling at 120 mph (constant the whole way for simplicity). This is a little bit simpler. I can calculate the change in gravitational potential energy for the fall and compare it to the kinetic energy of a dude going 120 mph.

Assuming the fall is at 120 mph, this would take about 40 seconds. The power for this case would be about 19,000 watts. Ok. I guess the stratos jump isn't too bad. Yes, it is more - but not way out of this range. So, maybe the jumper will get a little hotter - but he does have a space suit on.

One - one more thing

I asked the Red Bull Space jump guys for acceleration data when Felix actually does the jump, but I never officially heard back from them. Red Bull, if you read this - please?

rallain Wed, 05/19/2010 - 08:56

Tags

Google tells me that a joule is 0.239 calories, and one calorie will heat one gram of liquid water by 1 K. So if half of that energy goes to the jumper--lets call it 10^7 J for a 100 kg man. That's 2.39*10^6 calories going into 10^5 g of jumper, raising his average temperature from 37 C to 61 C if he's not wearing a space suit. (In reality, the leading edge of his body could get quite a bit hotter.) So he'd better be wearing that space suit, or else he will be toast.

How about a "space dive" from an orbiting vehicle? Say from the ISS.

At a mere 40 km, is the air really thin enough for a human body to hit Mach One? I would expect the wave drag to put a pretty serious limit on velocity at that point.

Secondly, I suspect that your 50/50 distribution of energy is pretty far off. First, for the near-sonic portions of the fall the compression heating of the air would be significant. Secondly, although the energy of impact between air molecules and the body would be distributed more or less evenly, there is also heat transfer from the body to the air -- and that air is cold. What's the wind chill of -40 C air at 300 m/s?

@D.C. Sessions,

I agree that maybe the 50/50 thing is a bit off. Look at a normal sky diver, they don't really get hot. It was just my first guess.

Re: Orbital Jump(#2)

Jumping from orbit is a little more complicated. You are moving at least 15,000 mph parallel to the ground. You will need a small rocket to change your orbit enough to intersect the atmosphere. When you hit the atmosphere, you will burn up unless you have a heat shield or tiles(i.e. shuttle) to convert all that velocity to heat that doesn't kill you.

An alternative would be a bigger rocket to dump all your orbital velocity, discard the rocket, and then fall with zero velocity relative to the ground. Would a small 3rd stage solid rocket give enough delta-V to kill your orbital velocity? Add in the control system also, unless manual control would be part of the challenge :-)

Tedd

It was just my first guess.

I would expect it to be a very good one from first principles. The problem is that there are other heat transfer processes which seem likely to dominate.

The good news is that it shouldn't be all that hard to calculate the loss rates. Fun assignment for students.

here is an article i just stumbled across today:

American Journal of Physics

"High-altitude free fall revised"

Jan Benackaa
Faculty of Natural Sciences, Constantine the Philosopher University, Tr. A Hlinku 1, SK-94974 Nitra,Slovakia
Received 24 September 2009; accepted 4 January 2010

@rob,

Thanks for the article - I am reading it now. How do I miss these things?

here is another one you might have missed twisted one 151 weblog. it has to do with heating of a skydiver. the coincidences abound today!

http://twistedone151.wordpress.com/2010/05/14/physics-friday-119/

May I ask what software you use to create the equation images?

@Swulf,

I use LaTeXit - http://chachatelier.fr/programmation/latexit_en.php

It is a small latex equation editor for mac. I then take screen shots of the equations. I know there are probably better ways, but this is quick for me. Also, you can drag vector graphic equations into Keynote for pictures.

Pumped up over triangles

rallain — Mon, 26 Apr 2010 07:40:11 +0000

Pumped up over triangles

I meant to mention this earlier since it happened a little while ago. There is this "mini-conference" with three schools: Southeastern Louisiana University, Southern Mississippi University, and the University of South Alabama. The purpose is to give students (and some faculty) a chance to present their work at a smaller conference. I really enjoy this, mostly because it is small and I get to see lots of undergrad talks. There are two talks that stuck in my head.

Dr. Jiu Ding "Dynamical Geometry: From Order to Chaos and Sierpinski Pedal Triangles"

Jiu Ding is a mathematics professor at Southern Mississippi. The talk was basically about inscribing triangles inside triangles using different "rules." The question was: what (if anything) do these triangles converge to as you keep on inscribing them. Here is an example for the paper: "Sierpinski Pedal Triangles", Jiu Ding, L. Richard Hitt, and Xin-Min Zhang.

I don't want to talk about triangles. I want to talk about how pumped up Jui Ding was about triangles. I mean PUMPED UP excited. I love that. You may ask: who cares about triangles? What's the big deal? The answer is: that doesn't matter. Are there any applications for triangles in triangles? Yes. But that is not what excites Jiu. He just likes seeing what will happen if you change the inscribing rules.

"What Goes Up Must Go Round: Final Analysis of the Autorotation of a Falling Maple
Seed"
by Ty McCleery and Dr. Lawrence Mead - USM Physics

Here is the abstract for this talk.

Seeds of maple trees increase their dispersion range by extending their fall time
through autorotation. By rotating, the seed remains airborne for a time longer than
without rotation, which increases its chances of being blown laterally by the wind and
therefore increases its dissemination. The seed's descent has been studied to determine
trends between flight characteristics and seed parameter's.

Theoretical and experimental analysis was used to determine what parameters control
the motion of the seed. These parameters were assumed to consist of the following: a
characteristic length of the seed (m), the seed's weight (N), the kinematic viscosity of air
(s/m), or the density of air (kg/m²).

Next, through dimensional analysis, these parameters were used to derive equations
for the dependent variables: terminal velocity, rotational speed, and coning angle. The
parameters and dependent variables were measured using experimental techniques such as
watching the falling seed's motion progressively with a still camera and a strobe light.

The measured dependent variables were then multiplied by powers of the parameters
in such a way that a dimensionless number was formed according to the Buckingham Pi
Theorem. The dimensionless numbers corresponding to each dependent variable (coning
angle, rotational speed, and terminal velocity) was plotted versus the others. Analysis of
these plots have yielded a relation between the parameters and the dependent variables.
The trends between the parameters and the dependent variables were also analyzed and
compared to previous research.

Again - here is an example of a cool project. I like this because:

It seems simple, but there are many layers to this question. Sort of a like an onion has layers. (not because it makes me cry)
There is not a lot of equipment or background work that needs to be done first. Really, anyone interested could just jump in and start working on this kind of thing.
There is not a clear motivation for the work other than "how does that work". That is not to say this won't lead to something useful, but that doesn't seem like the point.

The other undergraduate talks were also quite interesting, but I just wanted to give that as an example.

rallain Mon, 04/26/2010 - 03:40

Tags

Another example of a dangerous jump

rallain — Sat, 24 Apr 2010 14:13:53 +0000

Another example of a dangerous jump

Check this out.

So, the guy jumps from 150 feet into some cardboard boxes. Why are the boxes important? You want something that can stop you in the largest distance to make your acceleration the smallest. Here is my Dangerous Jumping Calculator. Basically, you put in how high you will jump from and how much distance you will take to land and it tells you your acceleration.

You will probably need this G-force tolerance info from wikipedia.

One problem - this calculator doesn't really work for this case. It doesn't take into account air resistance. Does air resistance even matter in this case? How about a plot of a person falling from 150 feet with and without air resistance? (I just used my code from the Red Bull Stratos Jump calculation)

This is a difference in speed of just about 3 m/s (or almost 7 mph). I guess every little bit helps.

rallain Sat, 04/24/2010 - 10:13

Tags

numerical calculation

python

spreadsheet physics

analysis

Your terminal speed will be about 200 km/h, which is 55 m/s. So after about 5 seconds it won't really matter how much more you're going to fall :)

The force the jumper feels is actually one G worse than your calculation. The jumper must feel os force of 1G just to counteract gravity, i.e. if the boxes gave him a force of 11G his speed would remain constant. So instead of feeling 14G, he is actually feel 15.

It is also pretty unlikely that the decleration is constant.

From your graph, the wind resistance slows his speed by about 10%. But that is a twenty percent drop in kinetic energy, so it does help considerably. Perhaps he could also try holding an umbrella as he falls as well, that would increase wind drag considerably.

Re: Omega Centauri at #2

As one nit-picker to another, I agree with what I think is your general idea, but I'm pretty sure when you say

> i.e. if the boxes gave him a force of 11G his speed would remain constant.

you're wrong.

Debris field for a broken meteor

rallain — Tue, 23 Mar 2010 17:42:42 +0000

Debris field for a broken meteor

I happened to catch two parts of two different episodes of Meteorite Men - a show about two guys that look for meteorites. In both of the snippets I saw, they were talking about a debris field for a meteor that breaks up. In these fields, the larger chunks of the meteorite are further down in the field. Why is this?

Let me approach this first from a terminal velocity view. This requires a model for air resistance. I will use the following:

Where:

rho is the density of air
A is the cross sectional area of the object
C is a drag coefficient that depends on the shape of the object
v is the speed of the object
And this gives a force with a direction opposite of the velocity vector

Let me assume that all the pieces of a meteor have the same density and shape - for simplicity, I will assume a sphere. Here is a diagram for two different sized pieces falling (straight down) at the same speed.

Meteor A (the big one) has a greater gravitational force because it has more mass. It also has a greater air resistance because it's cross sectional area is larger. I picked a speed so that meteor B would be at terminal velocity. This is when the air resistance has the same magnitude as the gravitational force. If I assume that meteor B has a radius of r_B and a density of rho_m then:

Where v_T is the terminal velocity. If I solve for this value, I get:

Here you can see the key point. The terminal velocity depends on the size. This is because the air resistance is proportional the area (r²) and the weight is proportional to the ~~area~~ volume (r³). These two things do not cancel.

Modeling a debris field

I have created a python model for shooting bullets. I can simply modify this to calculate the trajectory of a dozen or so different sized (but same shape and density) meteor pieces.

The following is a plot of the trajectory of a few pieces of a meteor. I (for random reasons) started the model at 5,000 meters above the ground moving at 350 m/s aimed 30 degrees below the horizontal. Here is what I get:

So, the bigger the piece, the farther it will go. My biggest piece was 1 meter.

rallain Tue, 03/23/2010 - 13:42

Tags

Hi Rhett,

I think that in the paragraph under your bottom most equation (the one the gives v sub T) you meant that weight is proportional to volume, not area.

I wonder what a realistic speed for a meteor is? I would imagine that they generally exceed the speed of sound so I'm not sure the naive equation for drag would apply. They typically burn so both the cross sectional area and volume would be decreasing as they descend.

@the other Rob,

Thanks for catching my mistake - I fixed it.

I really don't know about the speed of a meteor - I guess I could estimate this by modeling this as a rock coming from Jupiter or something. But, for this case, I just need the speed and height when it breaks up to get the debris field.

According to this site, a medium speed meteor has a speed of ~40 km/s. I remember reading (though I don't remember where) that large bolides are moving so fast that air doesn't have time to move around them and the huge pressure difference between the front side and the rear side is what can cause them to fragment.

I don't think your model of air resistance applies at all.

@Greg,

Yes - at 40 km/s, my model is fubared. But...the meteor will eventually slow down and then I can use the v^2 model of air resistance. The key is to consider when the thing breaks up. Either way, I assume the air resistance model for high speeds would still be proportional in some way to the cross sectional area and the weight is proportional to the volume.

Hello Rhett,

I am a proffessional meteorite hunter (semi-retired) and I know Mr. Notkin and Mr. Arnold (the Meteorite Men) very well. Your analysis is pretty much right on; I'm going to restate the facts in a way that may help crystalize the dynamic creation of strewnfield distribution.

There are two basic levels of velocity that matter, each with thier own set of physics; what we call the meteor's cosmic velocity (starting @ 72 to 12 km/sec), this velocity in a very general sense operates independent of gravity and air resistance and is referred to as a vector of that velocity. As the body transitions to terminal velocity then gravity and air resistance (winds aloft) become "everything" to the movement behavior of the body(s).

In 2003 when the Park Forest meteorite fell on Chicago, we learned just how much different the two modes of flight are and how counterintuitive a strewnfield distribution can seem w/o understanding them. We plotted the velocity vector expecting large stones farthest along the vector which was exactly what we found ... except the successively smaller bodies weren't on or along the vector(SSW-NNE) but extended out perpendicular to the vector (W-E) from the heavy impacts. The smallest were several kilometers away from that velocity vector. Why on earth (pardon the pun) was this distribution so skewed from the vel vector? The answer we found in the upper level winds; the jet stream was over Chicago that night and it's 150 mph west to east winds stretched the whole strewnfield to the east of the heavy stones. So there are two paradigms that follow the heaviest meteorite; they will be found furthest along the velocity vector AND closest to the velocity vector.

BTW all of this fluid and gravitational dynamics only helps when a body falls that is observed. If one happens to find a strewnfield that was not observed to fall, the only paradigms that help are the ellipsis rule and the general weight distribution. Strewnfields of any type tend to fall in an elliptical pattern.

PS: I almost forgot the third phase of strewnfield distribution TERRESTRIAL (erosional, wind, alluvial). Needless to say, this phase can obliterate any trace of its cosmic beginings if left to the job for long enough.

Take care and good physics!