Question 12: Dates and times trigonometry.

QUESTION 12.

(DATES AND TIMES TRIGONOMETRY).

11/2/1544 09:16:32 – 15/7/5792 15:31:54

T^1 = √(A/B) = 10^3 ms

a. What are the integers (k) of the years, months, days, hours, minutes and seconds between the two dates and times?

b. What are the values of s, m, h, d, M and y and what are the angles of their remainders or decimals in degrees and radians? Give sin(θ) and cos(θ).

c. What are the magnitudes of T^-1, T^2 and T^-2?

d. What are the values of t, A and B?

e. Reverse or undo t back into y/M/d/h/m/s integers (k) format.

f. Check that the remainder months add up to the remainder days.

Answer:

a.

k = y – M/(73/6) = 5792 – 1544 = 4248

k = M – d/30 = 7 – 2 = 5

k = d – h/24 = 15 – 11 = 4

k = h – m/60 = 15 – 9 = 6

k = m – s/60 = 31 – 16 = 15

k = s – cs/100 = 54 – 32 = 22

Because there is no obvious consensus on precisely how many days are in month, that is 28 – 31, therefore, when we are using months, in order to get the months and days to match and corroborate we must state that we are using 30 days in a month, therefore, we are using conversion factor of 73/6 months in a year. However, you could use a different whole number (a number without a decimal) for the conversion factor (or the amount of days in a month) such as 365/31.

Note: to convert a remainder decimal of a unit time into an angle of degrees and radians we times the decimal by 360 and 2π respectively.

For example:

s/60 = 11/30 × 360 = 132° Or s/60 = 11/30 × 2π = 11π/15

b.

s/60 = 22/60 = 11/30

θ = 132° or 11π/15

sin(θ) = 0.74314482547739

cos(θ) = -0.6691306063589

m = k + s/60 = 461/30

m/60 = (461/30)/60 = 461/1800

θ = 461/5° or 922π/5

sin(θ) = 0.99926291641062

cos(θ) = -0.0383878090875

h = k + m/60 = 11261/1800

h/24 = (11261/1800)/24 = 11261/43200

θ = 11261/120° or 11261π/21600

sin(θ) = 0.99775300871239

cos(θ) = -0.0669995045159

d = k + h/24 = 184061/43200

d/30 = (184061/43200)/30 = 184061/1296000

θ = 184061/3600° or 184061π/648000

sin(θ) = 0.77855054474696

cos(θ) = 0.62758190642674

M = k + d/30 = 6664061/1296000

M/(73/6) = (6664061/1296000)/(73/6) = 6664061/15768000

θ = 6664061/43800° or 6664061π/7884000

sin(θ) = 0.46719680793178

cos(θ) = -0.8841533479314

y = k + M/(73/6) = 66989128061/15768000

c.

T^-1 = √(B/A) = 10^-3 ks

T^2 = A/B = 10^6 ms²

T^-2 = B/A = 10^-6 ks²

d.

t = A/T^1 = y × 31536000 = 1.33978256122 × 10^11 s

X = 1.33978256122

A = tT^1 = X × 10^11 × 10^3 = X × 10^14 ms

B = A/T^2 = X × 10^14 / 10^6 = X × 10^8 ks

e.

y – M/(73/6) = t / 31536000 – M/(73/6) = 4248

M – d/30 = M/(73/6) × (73/6) – d/30 = 5

d – h/24 = d/30 × 30 – h/24 = 4

h – m/60 = h/24 × 24 – m/60 = 6

m – s/60 = m/60 × 60 – s/60 = 15

s – cs/100 = s/60 × 60 – cs/100 = 22

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – M/(73/6) = 133978256122 / 31536000 – M/(73/6) = 4248

M – d/30 = (4248.42263197615 – 4248) × (73/6) – d/30 = 5

d – h/24 = (5.14202237654839 – 5) × 30 – h/24 = 4

h – m/60 = (4.26067129645161 – 4) × 24 – m/60 = 6

m – s/60 = (6.25611111483865 – 6) × 60 – s/60 = 15

s – cs/100 = (15.3666668903188 – 15) × 60 – cs/100 = 22

Then insert the integers (k) into the y/M/d/h/m/s format:

t = 4248 years 5 months 4 days 06:15:22

f.

y – d/365 = t / 31536000×- d/365 = 4248

d – h/24 = d/365 × 365 – h/24 = 154

h – m/60 = h/24 × 24 – m/60 = 6

m – s/60 = m/60 × 60 – s/60 = 15

s – cs/100 = s/60 × 60 – cs/100 = 22

For example:

d = m × 30 + d

d = 5 × 30 + 4 = 154

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 133978256122 / 31536000 – d/365 = 4248

d – h/24 = (4248.42263197615 – 4248) × 365 – h/24 = 154

h – m/60 = (154.260671296452 – 154) × 24 – m/60 = 6

m – s/60 = (6.25611111483886 – 6) × 60 – s/60 = 15

s – cs/100 = (15.3666668903315 – 15) × 60 – cs/100 = 22

t = 4248 years 154 days 06:15:22

Hand written example:

5A13AA24-3926-4FF8-8BF0-65135CA0118A

Author:

Consider compassion, how can Australopithecus or Lower Palaeolithic man be compassionate toward animals, when they themselves were not yet masters of the animal kingdom or even worse still prey themselves? It is impossible, compassion simply did not exist. Compassion is technical, in that you must, for instance, first attain advanced weapons, technologies and infrastructure such as gunpowder, muskets, rifles, nuclear weapons, automobiles, militaries, police, emergency services, roads, buildings, bridges and skyscrapers etc before you can be compassionate toward animals. It is not a case of hey compassion for compassion’s sake like the Buddha. Compassion is not free of charge, it is a definite and tangible deal or bargain. Only now that I am invincibly safe and secure from wild animals in my city, town or fortress and surrounded by guns, and now that I have an overabundance and surplus of food, energy and resources etc can or will I be compassionate toward animals. It is like saying to ‘bear’ “I have a nuke now, therefore I am compassionate toward you.” This is something bear will never understand, in that it is ironic that once you attain nuclear weapons that you are therefore by definition compassionate toward animals. To reiterate, compassion is something technical, it is only attained through a collective effort, through taming the wild and through civilisation. You can only be compassionate once there is no competition.

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