Saturday, December 6, 2014

Some Science Stocking Stuffings


Einstein's Time Dilation

Recall that equation for time dilation is:

Δt = Δt0 / √(1 - u^2/c^2)

Where:
Δt = time observed by the person standing still
Δt0 = time observed by the traveler (in the observer's frame of reference)
u = speed of the moving object , which contains the traveler
c = speed of light in a vacuum = 299,792,458 m/s

In short, the traveler will note observe and note that time Δt0 has passed, which the person standing still observes that time Δt has passed. The closer someone goes to the speed of light, that less person she/he experiences. You will need to go super fast. Driving at 65 mph (29.0756 m/s) on the freeway won't cut it and here's why:

For 1 unit of time to pass for the person driving 65 mph (Δt0 = 1), the change of time for the observer (Δt) is:

Δt = 1/√(1 - 29.0756^2/299,792,458^2) ≈ 1.0000000000000047 (that is fourteen zeroes between the decimal point and the 4) (The Wolfram Alpha app was used for this calculation).

Virtually the same time passes.

Same deal regarding observing an airplane flying at 600 mph (268.224 m/s). For the person watching the plane, for a plane's passenger, pilot, or cocktail and peanut server to observe one unit of time, us watchers observe 1.0000000000004 (twelve zeroes) units of time.

If you want to be on an object where the people observing you experience twice the time (Δt = 2) you do (Δt0 = 1), you will need to travel (c*√3)/2 or 259,627,884.491 m/s (580,771,037.247 mph). That is 86% the speed of light!

Archer's Paradox (E.J. Rendtroff, 1913)

In archery, archers aim their arrows slightly to the side, instead of directly at the target. On the surface, that seems crazy. This is where the Archer's Paradox comes into effect.

Basically, when the arrow is shot, it travels in an "S" curve. Drawing a string makes the arrow bend such that he tip is pointed away from the target. As the arrow is fired and the string returns to the bow, the arrow bends the other way, turning the arrow back to the target.

Other factors to making accurate shots include the stiffness of the arrow, brace height, and wind conditions. I found this video by Billgsgate helpful:

http://youtu.be/bNlx6MBlymw

Galactic Coordinates

Galactic Coordinates are spherical coordinates that are set such as:

* The center is our sun.
* The coordinates are (l°, b°), where l is the longitude (0° to 360°, flat angle) and the latitude (-90° to 90°, height).
* Pointing "due east", l = 0° and b = 0° is pointing towards the center of the Milky Way. "Due west", l = 180° and b = 0° points away from the center of the Galaxy.

Using Wolfram Alpha ( http://m.wolframalpha.com ) and an online coordinator converter ( http://ned.ipac.caltech.edu/forms/calculator.html ), here are the 30° longitude markers when latitude is 0° (b = 0°):

* 0° points towards the constellation Sagittarius (as it should, it having the Milky Way center)
* 30° points towards Aquila the Eagle
* 60° points towards the Vulpecula the Fox
* 90° points towards Cygnus the Swan
* 120° points towards Cassiopeia the Vain Queen
* 150° points towards Perseus the Hero
* 180° points towards Auriga the Charioteer
* 210° points towards Monoceros the Unicorn
* 240° points towards Puppis (a ship's poop deck)
* 270° points towards Vela (a ship's sails)
* 300° points towards Crux, The Southern Cross
* 330° points towards Norma (a carpenter's square (measuring tool))

Spherical Lenses (see figure 2 below)

Variables:
P = place of object (with distance s)
P' = place of image (with distance s')
C = center of curvature
V = vertex of the lens

General relations:

tan α = h/(s - δ)
tan β = h/(s' - δ)
tan Φ = h/(R - δ)

However, if α < π/2 ( α < 45° ), we are dealing with paraxial rays:

1/S + 1/S' = 2/R

Surprisingly, the object is far from the vertex.


Forgive me if I repeat things. Instead of going out to clobber everyone last Black Friday for that extra 10% off, I stayed home and cracked open some books that I have been meaning to look at for months. I hope you find this enjoyable and insightful.

The best always,

Eddie


P.S. I was thinking about adding a section (or blog entry) about the age of Aquarius and some of the head scratchers I have about it. I am not sure if this subject would be appropriate for this blog. If you have any thoughts about this, or anything else, go and ahead and comment! - E.S.



This blog is property of Edward Shore. 2014

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