Video Analysis of Soyuz MS-10

There should be a grave yard for blog posts that start, but never get published.  Fortunately, I have this site.  Here I can share with you my failed posts.  Get ready.

It starts with this epic video from the Soyuz MS-10 failed launch.

That’s pretty awesome.  It’s doubly awesome that the astronauts survived.

Ok, so what is the blog post?  The idea is to use video analysis to track the angular size of stuff on the ground and from that get the vertical position of the rocket as a function of time.  It’s not completely trivial, but it’s fun.  Also, it’s a big news event, so I could get a little traffic boost from that.

How do you get the position data?  Here are the steps (along with some problems).

The key idea is the relationship between angular size, actual size and distance.  If the angular size is measured in radians (as it should be), the following is true L = r\theta where L is the length (actual length), theta is the angular size, and r is the distance.

Problem number 1 – find the actual distance of stuff on the ground.  This is sort of fun.  You can get snoop around with Google maps until you find stuff.  I started by googling the launch site.  The first place I found wasn’t it.  Then after some more searching, I found Gagarin’s Start.  That’s the place.  Oh, Google maps lets you measure the size of stuff.  Super useful.

Finding the angular size is a little bit more difficult.  I can use video analysis to mark the location of stuff (I use Tracker Video Analysis because it’s both free and awesome).  However, to get the angular distance between two points I need to know the angular field of view—the angular size of the whole camera view.  This usually depends on the camera, which  I don’t know.

How do you find the angular field of view for the camera?  One option is to start with a known distance and a known object. Suppose I start off with the base of the Soyuz rocket.  If I know the size of the bottom thruster and the distance to the thruster, I can calculate the correct angular size and use that value to scale the video.  But I don’t the exact location of the camera.  I could only guess.

As Yoda says, “there is another”.  OK, he was talking about another person that could become a Jedi (Leia)—but it’s the same idea here.  The other way to get position time data from some other source and then match that up to the position-time data from the angular size.  Oh, I’m in luck.  Here is another video.

This video shows the same launch from the side.  I can use normal video analysis in this case to get the position as a function of time.  I just need to scale the video in terms of size.  Assuming this site is legit, I have the dimensions of a Soyuz rocket.  Boom, that’s it (oh, I need to correct for the motion of the camera—but that’s not too difficult).  Here is the plot of vertical position as a function of time.sideviewsoyuzaccel.png

Yes, that does indeed look like a parabola—which indicates that it has a constant acceleration (at least for this first part of the flight).  The term in front of t2 is 1.73 m/s2 which is half of the acceleration.  This puts the launch acceleration at around 2.46 m/s2.  Oh, that’s not good.  Not nearly good enough.  I’m pretty sure a rocket has an acceleration of at least around 3 g’s—this isn’t even 1 g.  I’m not sure what went wrong.

OK, one problem won’t stop me.  Let’s just go to the other video and see what we can get.  Here is what the data looks like for a position of one object on the ground.

Data Tool 2018-11-04 13-31-41.png

You might not see the problem (but it sticks out when you are doing an analysis).  Notice the position stays at the same value for multiple time steps?  This is because the video was edited and exported to some non-native frame rate.  What happens is that you get repeating frames.  You can see this if you step through the video frame by frame.

It was at this point that I said “oh, forget it”.  Maybe it would turn out ok, but it was going to be a lot of work.  Not only would I still have to figure out the angular field of view for the camera, but I need to export the data for two points on the ground to a spreadsheet so that I can find the absolute distance between them (essentially using the magnitude of the vector from point A to point B). Oh, but that’s not all.  When the rocket gets high enough, the object I was using is too small to see.  I need to switch to a larger object.

Finally, as the rocket turns to enter low Earth orbit, it no longer points straight up.  The stuff in the camera is much farther away than the altitude of the rocket.

OK, that’s no excuse.  I should have kept calm and carried on.  But I bailed.  The Soyuz booster failure was quite some time ago and this video analysis wouldn’t really add much to the story.  It’s still a cool analysis—I’ve started it here so you can finish it for homework.

Also, you can see what happens when I kill a post (honestly, this doesn’t happen very often).

Actually, there is one other reason to not continue with this analysis.  I have another blog post that I’m working that deals with angular size (ok, I haven’t started it—but I promise I will).  That post will be much better and I didn’t want two angular size posts close together.

The end.

MacGyver Season 3 Episode 6 Hacks

Getting through a giant chunk of cement with acid.

Really, this isn’t a “Mac Hack” since MacGyver didn’t do it.  Instead it was someone else.  She used muriatic acid to help get through a tunnel that was plugged up with cement (or concrete—I always get those two confused).

But yes, muriatic acid will indeed “eat” through cement.  If the goal is to create a hole that will allow a human to get through, it’s not so bad.  You can use the acid to weaken enough of the structure that you can take it out in pieces.  You don’t have to dissolve all the cement.

Now for a bonus homework question—actually an estimation problem.  Suppose there is 10 feet of cement to get through.  If someone uses muriatic acid, how long would this hole take to make?  Go.

Thermite

Thermite is awesome—oh, and slightly dangerous.  Basically, it’s a chemical reaction between two metals in which one of the metals has the oxygen needed for the reaction (like iron oxide).  The key to getting this reaction to work is to have super tiny pieces of metal (like super super tiny).  Really, that’s the tough part.  But once you get that, the thermite gets really hot, really fast.  Hot enough to melt stuff.

Could it be used to close up gun ports in an armored vehicle?  Probably.

Pressure to open an armored vehicle

An armored car has armor.  That’s why they call it an armored car.  The primary role for the armor is to keep out things like bullets and people so that the stuff inside (probably money) is safe.

But what if you seal off the openings and then pump in some air?  The cool thing about air pressure is that even a small pressure increase can exert HUGE forces on a wall.  Let’s say that MacGyver doubles the atmospheric pressure inside the truck so that it goes from 10^5 \medspace \text{N/m}^2 to 2\times 10^5 \medspace \text{N/m}^2.  If you have a wall that has dimensions of 2 meters by 2 meters, there would be a net outward force of 4 \times 10^5 \medspace \text{N} pushing it outward.  For you imperials, that is like 90,000 pounds.

Oh, this is why submarines are cylindrical shaped.  Flat walls bend under great pressure.

DIY balance

I really don’t know how much detail to go over for this hack.  MacGyver builds a simple balance scale to find the weight of some money.  As long as the arms are equal, this should be easy.  Maybe in a future post (on Wired) I will show you a design for a symmetrical balance scale.

Pry Bar Lever

Again, there’s not too much to say here.  MacGyver uses a piece of metal to pry open a door.  This is why humans use crow bars.  In short, this is a torque problem.  The torque is the product of force and lever arm.  So a small force with a large lever arm (MacGyver pushing on the end of the bar) gives a small lever arm and huge force (from the tip in the door).

Ok, I lied.  Torque is way more complicated than that.  Really, it’s a vector product:

\vec{\tau} = \vec{F} \times \vec{r}

That’s a little better.

Oh, there were two good hacks that didn’t make it into the show.  Too bad.