Ideas

Relief map of Israel

http://commons.wikimedia.org/wiki/File%3AIsrael_relief_location_map-blank.jpg 

http://upload.wikimedia.org/wikipedia/commons/4/4e/Israel_relief_location_map-blank.jpg 

Show the locations of Ramah, Hazor, and where Sisera and Barak meet.

Helium

From rebuilding adelle.

Expansion ratio of Helium. 1:757.  Density of Helium at STP  0.1786 g/L.  1.5 Kg / 0.0002786 Kg/L = 8398 L.  8398 L / 757 = 11 L = 0.011 m3.   is a cube 22 cm/side.

Period of an orbit

From http://answers.yahoo.com .

v**2 = Gm/r and r = Gm/v**2

G is the gravitational constant, 6.67 x 10*-11

r = 7.0E+5 m + 63.71E+5 m = 7E+6 m r=7.0E+5 + 6371E+3

T = sqrt(4 * math.pi )

From Wikipedia:

P = 2

Where a is the semi-major distance, and μ is the standard gravitational parameter, GM.  G is 6.67384E-11 m3/(kg*s2).  μ for the earth (mass 5.972E+24 kg ) is 3.985617248e+14.  a=500 Km+ Rearth (6,371 Km) = 6871 Km =6.871E+6 m

In [1]: G=6.67384E-11

In [2]: Me=5.972E+24

In [3]: u = G * Me

In [4]: print u

3.985617248e+14

In [5]: a = 6.871E+6

In [6]: import math

In [8]: print math.sqrt(a**3/u)*2*math.pi

5668.41966922

In [9]: print math.sqrt(a**3/u)*2*math.pi/3600.

1.57456101923

gravitational constant = 6.67384 × 10-11 m3 kg-1 s-2

More about calculator.

Mass of the earth 5.972E+24 kg

6.67E-11 m**3/(Kg*s**2)*



r = 6.67 x 10*-11(6 x 10*24)/8000*2

r = 6.25 x 10*6 m

Kepler's third law gives us the period

For circular orbits, Kepler's 3rd Law is also commonly represented as

Where  is the period,  is the Gravitational constant,  is the mass of the larger body, and  is the distance between the centers of mass of the two bodies.

T = math.sqrt( 4*math.pi**2*r**3/(6.67E-11*5.972E+24))

# Mass of Europa is 4.7998×1022 kg

# Mass of the Earth is 5.972E24 kg

# Radius of the earth is 6371000 m

import math

# M = 4.7998E22 

M = 5.972E24

# r = 200000 + 1560800

r_earth = 6371000

r_orbit = 500000

r = r_earth + r_orbit

G = 6.67E-11

T=math.sqrt(4*math.pi**2*r**3/(G*M))

print “Orbital period is %f seconds, %f hours” % (T, T/3600.)

Circumference = r *2 * math.pi

V = Circumference / T

print “The velocity is %f meters/sec or %f kilometers/hour” % (V, V*3600./1000.)

Orbital period is 8204.845830 seconds, 2.279124 hours

The velocity is 1348.402264 meters/sec or 4854.248149 kilometers/hour

Degrees Celsius °C

electroenchephalographically

The recoil of a laser

What is the recoil of a megawatt laser.  Assume a Neodymium YAG laser with a wavelength of  1064 nm.

See http://en.wikipedia.org/wiki/Photon 

The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):

where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.[16]

Since p points in the direction of the photon's propagation, the magnitude of the momentum is

E =

h = Planck’s constant = 6.62606957(29)×10−34 J*s  

hc = 1.98644568×10−25 J*m

http://en.wikipedia.org/wiki/Planck_constant

c = speed of light =  299 792 458 m / s = 3.0E+8m/sec

E =  = 1.86*10-37J

A megawatt-second = 1 megaJoule  1 MJ = E*n  n = 1MJ/E = 1.0E+6 J /1.86E-37 J/photon = 5.37E+42 photons

Kg*m/s = 6.62E-34/1064E-9 = 6.23E-28 N*s/photo

6.23E-28 N*s/photon * 5.37E+42 photons = 3.34E+15 N*s

http://en.wikipedia.org/wiki/Radiation_pressure

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/Pressure = Intensity / c  Intensity is W/m2 .  Pressure is Force/area = N/m2 .  P = F/A = I/c  F=power/c

1.0E+6W/3.0E+8 m/s = .003 W*s/m = .003 N

units

You have: (1.0E+6 watts)/(3.0E+8 meters/sec)

You want: newtons

        * 0.0033333333

        / 300

You have:

Degrees Celsius ℃ U+2103 Latin Compatibility row 5

Ship’s atmosphere: http://en.wikipedia.org/wiki/Atmospheric_pressure 50 Kilopascals  (Standard atmosphere is 101 Kilopascals.  Oxygen is 40%, Nitrogen is 60%.  That causes problems with the bends.  The exploding brick.  The boiling point of Nitrogen.


Deborah’s chronology

http://www.bible.ca/archeology/bible-archeology-exodus-route-date-chronology-of-judges.htm

This disagrees with

First Battle with Unrar, 1119 BCE.  Second Battle with Unrar 2456 CE.

Event

Date

Deborah’s Age

Lapidot’s age

Achiya’s age

Barak, son of Achiya

Born

March 1991

0

Planned graduation:

May 2013

22

Begins training

2372 CE

22

Completes training

2374 CE

24

Arrives in Canaan

1126 BCE

24

Lapidot born

1125 BCE

25

0

Marries Lapidot

1119 BCE

40

15

Achiya born

1119 BCE

41

16

0

Returns from Europa

1114 BCE

46

22

5

Defeats Siseria

1113 BCE

47

23

6

Achiya marries Hodoya

1102 BCE

58

34

17

Barak, son of Achiya born

1101 BCE

59

35

18

0

Lapidot dies

1092 BCE

68

44

27

9

Achiya dies

1078 BCE

82

41

23

Returns to the future

1073 BCE
2456 CE

87

28

Walter returns to his time

2374 CE

87

Returns to the present

May 2013

88