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Welcome to 2021, and may 2020 stay (far) behind us!
Some fun things to occupy your time:
- Find the center of a circle
- Make some quiche
- Enumerate permutations of things
- Have some laughs here, here, here, here, or here.
This morning, July 20, 2021, Blue Origin’s New Shepard rocket launched in to space for its first 11-minute passenger flight.
Congrats to Bezos and team!
HOWEVER, the commentator on one of the science channels made this statement: “The rocket needs to go high enough to reach orbit, even briefly”
A popular misconception is that orbit and outer space are the same thing – actually, I’ve written about this previously. However, I would expect a science commentator on a science-related channel to know the difference!
Just to recap:
Even hundreds of thousands of miles away from the surface, any object is still within the Earth’s gravitational influence. So, if you fire a rocket straight up (perpendicular to the Earth’s surface), even hundreds or even thousands of miles in to space, eventually, the rocket will run out of fuel, and the Earth’s gravity will pull it back down.
In orbit, however, an object’s velocity parallel to the Earth’s surface creates centrifugal force which balances the pull gravity.
E. A rocket launched perpendicular to the Earth’s surface eventually runs out of fuel and is pulled back down by gravity. F. To achieve orbit, a rocket must attain both height and velocity parallel to the surface.
My understanding of New Shepard is that it will basically travel straight up, give the passengers a few minutes of weightlessness, and then return to Earth.
Although gravity will be acting on the spaceship and its passengers during the descent, the passengers will experience weightlessness (which is different than outer space and also different than orbit) because gravity acts on both nearly-equally. The state or sensation of weightlessness occurs when there are no net forces acting on the passengers, which is only true inside the ship, and only relative to the ship itself.
Although the exact height of the New Shepard’s mission isn’t listed, it is expected to go “well above” the Kármán Line. So if we were to guess 65 miles, we would be in the right ball park.
This means that New Shepard’s velocity at 65 miles above the Earth’s surface will be zero, after its rockets turn off, and gravity bleeds away all of its upward velocity. At that exact point, with gravity acting constantly, it will begin to accelerate back down to Earth. Until the ship hits turbulence in the atmosphere, or they fire their re-entry rocket, the passengers will continue to fall at an ever faster rate toward Earth, pulled by gravity, yet experiencing weightlessness within the ship.
In contrast, to attain orbit at 65 miles above the Earth’s surface, the ship would have to be travelling almost 17,780 miles per hour parallel to the Earth’s surface in order to have enough centrifugal force to exactly counter the force of gravity.
How do we go about calculating this? Glad you asked…
Force of gravity:
F = m * g
- m = mass of the object
- g = acceleration due to gravity
Centrifugal force (same as centripetal force, but in the opposite direction):
F = m * v^2 / r
- m = mass of the object
- v = linear velocity
- r = radius of the orbit (from the center of the Earth)
In orbit, the net force is zero, so:
m * g = m * v^2 / r
We quickly see that mass is irrelevant, and we can solve for v:
v = sqrt( g * r )
- sqrt = square root function
If we measure gravity and orbital radius in feet, we get orbital velocity in feet per second. To simplify things, we can use 78,545 miles per hour^2 as the acceleration due to gravity:
v = sqrt( 78545 * r )
The radius of the Earth is about 3960 miles. To find orbital velocity, we have to have the radius from the center of the Earth, which include’s the Earth’s radius plus the height above the surface:
v = sqrt( 78545 * ( 3960 + r ) )
If we plug in 65 miles, we get about 17,780 miles per hour required for orbit.
So again, good luck to Bezos and team – although you won’t be “in orbit”, you will definitely be in outer space.
Previously, I’ve written about gun-related movie myths in “Movie Myths: Guns – Part 1“, and in there, I described in detail how guns work – both automatics and revolvers.
Since that time, I’ve thought about making a state diagram that helps explain some of the common gun-related continuity errors that you regularly see in movies and TV.
The result is the diagram that you will see in this article, along with a higher-resolution PNG and PDF that you can download for free, which would make a nice wall decoration for any gun enthusiast.
Download the PDF here, or the PNG here.
Read on for more information about state diagrams and gun-related movie continuity errors…
I bought a wood lathe recently, and the first time I tried to center some wood, I went through the painful process of remembering how to find the center of a circle. This, despite having had three years of mechanical drawing classes.
Starting with a pencil and straight edge:
The center of the circle is at the intersection of the vertical (E) and horizontal (H) center lines.
Let’s call the distance from the corner of the straight edge to its mark S.
Let’s call the points where the horizontal center line meets the circle’s edges H1 (left) and H2 (right), and the point where the vertical center line meets the bottom V.
If we start at M and walk clockwise around the circle, we would pass through the points as follows: M (start) -> H2 -> V -> H1 –> M (back to start).
The circle’s radius, R, is 1/2 the distance of M -> V or H1 -> H2. The distance from M -> H2 would be SQUARE_ROOT( R^2 + R^2), and the same is true for H2 -> V, V -> H1, and H1 -> M.
If we assume that the radius R is “1 unit” (the actual length is irrelevant), then SQUARE_ROOT(1^2 + 1^2) = SQUARE_ROOT(2), or about 1.41.
If the radius R = “1 unit” then the diameter = 2 * R = “2 units”.
By selecting S such that S is 3/4 of the diameter, 3/4 * 2 = 1.5. Since S>1.41, and we make two consecutive S-length marks around the circle, we know for a fact that 2*S > (M -> H2 -> V).
By doing this twice, once clockwise, and once counterclockwise, we get lines C and D. We can call its intersection point T.
It’s easy to prove that this forms two congruent triangles: (M->A-> T) and (M->B->T). If the line M -> T is the base of both, then because they are congruent, the height of both are the same, which means that M -> T perfectly bisects (A -> B).
Since this line (M -> T) also bisects the equilateral triangle M -> A -> B which is inscribed within the circle, that line must pass through the center of the circle as well.
Carrying M ->T out to the edge of the circle gives us point V.
By drawing a line from V of length S to the edge, we get line F, which crosses line A. We can call this intersection point U.
If we knew where the center of the circle (point O) was located, we could make the congruent triangle argument, that M -> U -> O is congruent to V -> U -> O, and just as with the vertical center line bisecting M->A->B, U -> O would perfectly bisect the larger equilateral triangle M -> U -> V.
On the other side, we draw line G of length S, which crosses line B at point W. Same congruency argument: M -> W -> O is congruent to V -> W -> O, and W -> O bisects equilateral triangle M -> W -> V.
Because M -> U = V -> U we know that M -> U -> V is equilateral, and the same for M -> W -> V. Further, because of the congruency argument, we know that M -> U = M –> W and V -> U = V -> W, and therefore (M -> U -> V) is congruent to (M -> W -> V).
If line U-> O bisects triangle M->U->V, then it must intersect M->V exactly at its center. Same for W->O and triangle M->W->V.
Therefore, W->U passes through the center of both triangles at point O, where W->U intersects M->V.
And therefore, point O must be at the center of the circle, because V->O = M->O, and we know that M->V passes through the center of the circle.
Note that this trick ONLY works if the the mark on the ruler (distance S) is > 1.41 times the radius.
I make quiche once or twice per year, and I don’t have a specific recipe that I follow – I usually google for things like portions, temperature, and cooking time.
This year, I decided to take some notes and provide a general quiche framework that will work for any kind of quiche you want.
The goal is to have 4 cups of eggs, cheese, plus other stuff (filler ingredients) that we pour in to a 9″ pie crust. Although a 9″ pie crust will hold about 5 cups, the quiche will rise during the cooking process.
One major tweak from “normal” quiche recipes: Instead of milk, I use sour cream, which has much more fat in it and also significantly improves the flavor. Because sour cream is more dense, I use bread (chopped up) so that you don’t end up with a 2″ thick egg-crete block.
Let me start off by saying “I HATE FTP!”
It was developed in the early 1970’s, before the Internet was really the Internet, and before the existence of firewalls or any other type of network security. Over the last nearly-50 years, it has picked up so many band-aids that the entire protocol is now just basically one big kludge.
Despite its shortcomings, and despite the availability of newer and far superior file transfer tools, banks, governments, and anyone else with a mainframe continues to use it.
So, unfortunately, people like you and I get stuck supporting it.
If you’re having an FTP problem, or are just morbidly curious to learn about an antiquated protocol, read on…
