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Footnotes for the 8 UFO Chapters
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Chapter: 11.14
(Section 11: Understanding God and His Universe)
Copyright © Michael Bronson 1997, 1999, and 2000
BibleHelp.org
Summary
Since the 8 chapters on UFOs have many calculations in the footnotes, I created a separate web page to put these footnote calculations.
Click Here for a book version of this material
Footnote 2.1
Magnification of part of our galaxy: Pictures 2 and 3 are a magnification of our galaxy (Milky Way). Picture 1 is not of our galaxy. Obviously, since there’s no way of getting a photograph of our galaxy, I used a photograph of a galaxy that looks like ours (NGC 253 from the Sculptor Constellation).
Footnote 2.2
The formula used to calculate the travel time from the closest star:
A light year is the distance light can travel in a year
Light travels at 186,282 miles per second
There are 31,536,000 seconds in a year
That means light travels 5.87 trillion miles in one year
The closest star is 4.2 light years away
Therefore, the closest star is 24.7 trillion miles away
Flying at 340,000 miles per hour, it would take 8,284 years to travel this distance (24.67 trillion miles / 340,000)
Footnote 3.1
To say that there is "no gravity in space" is inaccurate. There is no limit to the influence of gravity. The gravitational pull of a galaxy can be felt hundred of millions of light years away. As a general rule, small objects in outer space appear not to be affected by gravity in space. It should be noted that objects orbiting Earth are still slightly affected by Earth’s gravitational pull. However, the speed of the spacecraft is timed in such a way as to offset the effects of gravity.
Footnote 3.2
This is the formula to calculate the energy needed to propel a spaceship to 50% of the speed of light. This does not include the energy needed to get the spaceship out of earth’s orbit. The calculations are for a spaceship that is already out of our solar system.
The weight of the spaceship (used in these calculations) is the normal weight of NASA’s space shuttle when it goes on a one-week trip. These calculations do not include the weight of any of the extra supplies needed for the 8.2 years to get to the nearest star. These calculations don’t even include the energy needed to move the engine and fuel that would be necessary to 50% of the speed of light. It only accounts for the energy needed to get the ship itself to 50% of the speed of light.
The formula below is to calculate getting one pound to 50% of the speed of light.
50% of speed of light = 1.5 x 108
1 pound = 0.4536 kg
Kinetic energy = (.5) (mass) (velocity) (velocity)
Kinetic energy = (.5) (.4536 kg) (1.5 x 108) (1.5 x 108)
Kinetic energy = (5.1 x 1015)
The formula below compares this energy to an atomic bomb
Energy of Hiroshima atomic bomb = 5.2 x 1013
Kinetic energy needed to propel the ship = (5.1 x 1015)
Number of atomic bombs = (5.1 x 1015) / (5.2 x 1013)
Number of atomic bombs = 97.998
The calculations to find the energy to propel the whole ship
Spaceship weighs 231,053 pounds
Amount of energy to propel whole ship = (energy for 1 pound) (weight of ship)
Amount of energy to propel whole ship = (5.1 x 1015) (231,053)
Amount of energy to propel whole ship = 3.81 x 1021
The formula below compares the energy to propel the whole spaceship to the total energy used in the U.S. for an entire year.
Amount of energy consumed in US in a year = 3,015,383 million kilowatt hours
1 kilowatt hour = 3,600,000 joules
3,015,383 million kilowatt hours = 1.09 x 1019 joules
Number of years worth of energy = (energy for ship) / (energy used by US)
Number of years worth of energy = (1.18 x 1021) / (1.09 x 1019 joules)
Number of years worth of energy = 108.464 years
Footnote 3.3
This is the formula to calculate the energy needed to propel a spaceship to 90% of the speed of light. This does not include the energy needed to get the spaceship out of earth’s orbit. The calculations are for a spaceship that is already out of our solar system.
The weight of the spaceship (used in these calculations) is the normal weight of the space shuttle when it goes on a one-week trip. These calculations do not include the weight of any of the extra supplies needed for the 5 years to get to the nearest star. These calculations don’t even include the energy needed to move the engine and fuel that woule be necessary to propel it to 50% of the speed of light. It only accounts for the energy needed to get the ship itself to 50% of the speed of light.
The formula below is to calculate getting one pound to 90% of the speed of light.
90% of speed of light = 2.7 x 108
1 pound = 0.4536 kg
Kinetic energy = (.5) (mass) (velocity) (velocity)
Kinetic energy = (.5) (.4536 kg) (2.7 x 108) (2.7 x 108)
Kinetic energy = (1.65 x 1016)
The formula below compares this energy to an atomic bomb
Energy of Hiroshima atomic bomb = 5.2 x 1013
Kinetic energy needed to propel the ship = (1.65 x 1016)
Number of atomic bombs = (1.65 x 1016) / (5.2 x 1013)
Number of atomic bombs = 317.515
The calculations to find the energy to propel the whole ship
Spaceship weighs 231,053 pounds
Amount of energy to propel the whole ship = (energy for 1 pound) (weight of ship)
Amount of energy to propel the whole ship = (1.65 x 1016) (231,053)
Amount of energy to propel the whole ship = 3.81 x 1021
The formula below compares this energy to the whole spaceship to the total energy used in the U.S. for an entire year.
Amount of energy consumed in US in a year = 3,015,383 million kilowatt hours
1 kilowatt hour = 3,600,000 joules
3,015,383 million kilowatt hours = 1.09 x 1019 joules
Number of years worth of energy = (energy for ship) / (energy used by US)
Number of years worth of energy = (3.81 x 1021) / (1.09 x 1019 joules)
Number of years worth of energy = 351.426 years
Footnote 4.1
Source of paint flake hitting the Challenger:
Discover Magazine, April 1990
U.S. News and World Report, October 22, 1990
-Artificial Space Debris, Johnson and McKnight, p 69
Footnote 4.2
To find the impact power of an object, we use the following formula
Kinetic energy = (.5) (mass) (velocity) (velocity)
o Where energy is measured in Joules
o Where mass is measured in kilograms
o Where velocity is measured in meters per second
Footnote 4.3
Determining the depth of penetration of a high-speed projectile is very complicated. Once you have calculated the kinetic energy generated from the impact, you need to know several things about the material being impacted. The material’s density, its "elasticity," its heat of fusion, the speed in which sound travels through it, etc.
Density is the amount of matter packed into a certain amount of space. For example, a square inch of gold is much more dense than a square inch of wood. Density is an important factor in calculating the depth of penetration. Let’s say we have two objects that weigh the same (have the same mass), but one is much more dense than the other. In an impact, the less dense object would make a wider and shallower hole while the denser object would make a narrower and deeper hole.
The outer shell of most spaceships is made out of aluminum (with a density of 2.7 g/cm3). The density of potential object hitting the spaceship ranges from .05 g/cm3to 8 g/cm3. To simplify the calculations, scientists often use a density somewhere between these two extremes (impacts of aluminum on aluminum). The Bjork’s equation for this type of impact is:
Penetration = 1.09(mass x velocity)1/3
Penetration in cm
Mass in grams
Velocity in meters per second
Although this formula will give you a ballpark idea of what happens with the "slower" high-speed impacts, it should not be used on impacts higher than 10 km/sec. We know very little about ultra high-speed impacts. We do know that with speeds between 3 and 6 km/sec, the impact site is treated as a solid. At impact speeds around 20 km/sec, the impact site is treated as a liquid and higher speeds it is treated as a vapor.
Our knowledge of hypervelocity impacts comes from experimenting with hypervelocity guns. Shooting various objects at various speeds has given us a wealth of information. Since these guns are currently limited to about 7 km/sec (4.4 miles/sec) computer models are used to simulate higher speeds. Of course, these computer models are not very accurate on speeds past 20 km/sec.
For those who are interested, I have listed 3 other formulas used to calculate penetration depths. Since these formulas are beyond the scope of this chapter, I am not including any explanation for them.
Whipple’s equation: P = K1 (1/ pi Qع)1/3 E 1/3
P = |
Penetration depth in a thick target material |
K1 |
Constant of proportionality |
E |
Meteorite (debris) energy |
Q |
Density of target material |
ع |
Heat of fusion of target material |
Kornhauser’s equation: h= K2(T/E)1/3 (E/EO)0.009
h |
Penetration (depth of crater) |
K2 |
Constant of proportionality |
T |
Kinetic energy of projectile |
E |
Modulus of elasticity of target |
EO |
Reference modulus |
Summer’s equation: P/d = 2.28(QP/QT)2/3 (Vc)2/3
P |
Penetration depth in a thick target material |
d |
Diameter of projectile |
QP |
Density of projectile |
QT |
Density of target |
V |
Projectile velocity |
c |
Speed of sound in target material (5.1 x 105 cm/s for aluminum) |
Footnote 4.4
The atomic bomb dropped on Hiroshima created 5.2 x 1013 Joules of energy. When the spaceship (flying at half the speed of light) hits the object the size of a pea, it creates 3.65 x 1014 joules of energy. Therefore, the impact creates the kinetic energy of 7.03 atomic bombs
1 Ton of TNT = 4.184 x 109 Joules
1 Pound of TNT = 2,092,000 Joules
1 Hand grenade = 150 grams of TNT = 690,360 Joules
1 bullet (30 Cal) = 464 Joules
Footnote 5.1
To provide round-the-clock force field protection for the spaceship (traveling 90%) we need to find the number of seconds in 5 years.
Seconds in 5 years = (sec) (min) (hours) (days) (years)
Seconds in 5 years = (60 sec) (60 min) ((24 hours) ((365 days) (5 years)
Seconds in 5 years = 157,680,000 seconds
Footnote 5.2
The aluminum object the size of a pea has the mass of 10 grams. The aluminum object the size of a softball has a mass of 10,000 grams. Therefore, its impact would be 1,000 times greater
Footnote 6.1
These calculations will shrink the trip to our nearest star to a trip similar to the distances across the United States. This calculation is assuming that these is only one object near the path of the spaceship once every million miles.
Distances to nearest star = 24.7 trillion miles
# of objects in its pathway = (24.7 trillion miles) (1 object per million miles)
# of objects in its pathway = 24.7 million objects
# of objects across the US = 3,000 miles
# of object per miles = (# of objects) / (3,000 miles)
# of object per miles = 24.7 million) / (3,000 miles)
# of object per miles = 8233
# of object per foot = 1.6
Footnote 7.1
A spaceship is traveling at 50% of the speed of light (93,000 miles per second) and there is a boulder 3,333 miles away. The formula below calculates the time it will take to reach the boulder.
Time = distance / speed
Time = 3,333 miles / 93,000 miles per second
Time = .03584 seconds
Footnote 7.2
At present, we can see about 13 billion light years in all directions. Therefore, the "known" universe is at least 26 billion light years wide.
Item A: Light travels at 186,282 miles per second
Item B: A light year is 5.874 trillion miles
Item C: 26 billion light years would be 1.527 x 1023
Item D: The Earth’s diameter is 7,926 miles
The relative miles-per-second is calculated by: (Item A) (Item D) / (Item C)
This gives us: 9.666 x 10-15 miles per second
(9.666 x 10-15) (60 seconds) (60 minutes) (24 hours) (365 days)
This gives us 3.048 x 10-7 miles per year
There are 63360 inches per mile (5280 feet x 12 inches)
(3.048 x 10-7) (63360) = .0193 inches per year
Footnote 7.3
A spaceship is sending out a scanning signal to look for objects in its path. This signal is so narrow and highly focused that it is only scanning a piece of sky that is the size of a grain of sand (As described in the chapter). After 10,000 miles, this narrow signal will have expanded a great deal. This footnote calculates the area of growth. This area of growth (referred to as the "danger zone") was calculated by using trigonometry on a triangle:
Angle C = less than 8/100th of a degree (.078 degree)
Side a = 10,000 miles
Side b = 10,000 miles
Side c = ?
The formula to find side c is: c2 = a2 + b2 – 2ab(Cos C)
c2 = (10,000) 2 +(10,000) 2 –2(10,000)(10,000)(cos .078)
c2 = (200,000,000) – (200,000,000)(.999999086)
c2 = 185.329
c = 13.614 miles
Angle C (used in the above formula) was derived by the following formula:
The grain of sand is approximately 1 mm
The grain of sand is viewed at approximately 740 mm away
The following triangle is used for the equation
[pic of triangle]
(cos C) = c2 - (a2 +b2) / -2ab
(cos C) = 12 - (7402 +7402) / -2(740)(740)
(cos C) = 1 - (1,095,200) / (-1095,200)
angle C =.0077426 degrees
Footnote 8.1
A spaceship is traveling at 50% speed of light and detects a boulder. It has .0358 seconds to make an adjustment of seven miles (36,400 feet) to avoid hitting the boulder. The following formula is used to determine the G-force exerted on the spaceship:
acceleration = 2(distance) / (time)(time)
acceleration = 2(36,400 feet) / (.0358 seconds)(.0358 seconds)
acceleration = 56,802,222 feet/second2
One G-force is equal to an acceleration of 32 feet per second2. Therefore:
G-force = (56,802,222) / (32)
G-force = 1,775,069
Footnote 8.2
A spaceship is traveling at 50% speed of light and detects a boulder. It needs to make an adjustment of seven miles (36,400 feet) to avoid hitting the boulder. The following formula is used to determine the response distance needed to keep the G-force at 3 Gs:
1 G = an acceleration of 32 feet per second2, therefore,
3 Gs = an acceleration of 96 feet per second2
Time = square root of [2(distance) / (acceleration) ]
Time = square root of [2(36,400 feet) / (96 feet per second2)]
Time = 27.5 seconds
The spaceship is traveling at 93,000 miles per second
Distance = (27.5 seconds)(93,000 miles per second)
Distance = 2,561,020 miles
The above distance is the response distance (the distance needed to make a safe adjustment. Now, we need to find the total distance (the distance when the signal is first sent out). We already know (from the chapter) that the total distance is 3 times the response distance(3,333 miles out of 10,000 miles). Therefore, the total distance is 7,683,060 miles.
Footnote 8.3
A spaceship is traveling at 90% speed of light and detects a boulder. It has .0028285 seconds to make an adjustment of seven miles (36,400 feet) to avoid hitting the boulder. The following formula is used to determine the G-force exerted on the spaceship:
Acceleration = 2(distance) / (time)(time)
Acceleration = 2(36,400 feet) / (.0028285 seconds)(.0028285 seconds)
Acceleration =72,800 / 000008 seconds2
Acceleration = 9,099,531,090 seconds2
One G-force is equal to an acceleration of 32 feet per second2. Therefore:
G-force = (9,099,531,090) / (32)
G-force = 284,360,346
Footnote 8.4
A spaceship is traveling at 90% speed of light and detects a boulder. It needs to make an adjustment of seven miles (36,400 feet) to avoid hitting the boulder. The following formula is used to determine the amount of time needed and distance to keep the G-force at 3 Gs:
1 G = an acceleration of 32 feet per second2, therefore,
3 Gs = an acceleration of 96 feet per second2
Time = square root of [2(distance) / (acceleration) ]
Time = square root of [2(36,400 feet) / (96 feet per second2)]
Time = 27.5 seconds
The spaceship is traveling at 167,653 miles per second
Distance = (27.5 seconds)(167,653 miles per second)
Distance = 4,616,803 miles
The above distance is the response distance (the distance needed to make a safe adjustment. Now, we need to find the total distance (the distance when the signal is first sent out). We already know (from the chapter) that the total distance is 21 times the response distance. Therefore, the total distance is 96,952,876 miles.
Footnote 8.5
A spaceship is traveling at 50% speed of light and detects a boulder. It has .0358 seconds to make an adjustment of 30 feet to avoid hitting the boulder. The following formula is used to determine the G-force exerted on the spaceship:
acceleration = 2(distance) / (time)(time)
acceleration = 2(30 feet) / (.0358 seconds)(.0358 seconds)
acceleration = 46,815 feet/second2
One G-force is equal to an acceleration of 32 feet per second2. Therefore:
G-force = (46,815) / (32)
G-force = 1,463
Footnote 8.6
A spaceship is traveling at 90% speed of light and detects a boulder. It has .0028285 seconds to make an adjustment of 30 feet to avoid the boulder. The following formula is used to determine the G-force exerted on the spaceship:
acceleration = 2(distance) / (time)(time)
acceleration = 2(30 feet) / (.0028285 seconds)(. 0028285 seconds)
acceleration = 7,499,614 seconds2
One G-force is equal to an acceleration of 32 feet per second2. Therefore:
G-force = (7,499,614) / (32)
G-force = 234,363
Conversion Table:
1 inch = 2.54 centimeters (cm) 1 mile = 1.609 Kilometers (km) 1 mile = 1,609.344 meters (m) 1 kilometer = 0.621371 miles 1 meter = 3.28084 feet 1 meter = 1.093613 yards 1 foot = 0.3048 meters 1 pound = 0.4536 kilograms (kg) 1 pound = 453.6 grams (g) 1 kilogram = 2.204586 pounds 1 gram = 0.002204586 pounds Speed of light = 186,282 miles per second Speed of light = 299,791,819 meters per second |
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