Klingons Break Treaty

[wwjd_kilden]

hmmmmmmmmm.....


I prefer my days to be 14-16 hours long and my nights to be 10 hours long

 

[resurrection33]

This is Klingon Red Leader: Have been severely wounded, shall not be fighting for a while. Spock shall be taking over the regime. It was good fighting alongside and against you blokes, but I must retire. This was Klingon Red Leader, over and out. *salutes with a solemn face*

I hope you get well soon, Klingon Red Leader.

 

[resurrection33]

Metric'DAY''''months''''days'''''hours''''minutes'''''''seconds
'1,000.000'10,957.50''365.25''8,766.00'525,960.00'31,557,600.00
'''100.000''1,095.75'''36.53''''876.60''52,596.00''3,155,760.00
''''10.000''''109.58''''3.65'''''87.66'''5,259.60''''315,576.00
'''''1.000'''''10.96''''0.37''''''8.77'''''525.96''''''31,557.60
'''''0.100''''''1.10''''0.04''''''0.88''''''52.60''''''3,155.76
'''''0.010''''''0.11''''0.00''''''0.09'''''''5.26''''''''315.58
'''''0.001''''''0.01''''0.00''''''0.01'''''''0.53'''''''''31.56

so 1000 metric days in a year

8.77 OLD hours in a metric day

So we can work eight hours a day and have .77 hours to do whatever we want. Get to work, everyone!

 

[resurrection33]

hmmmmmmmmm.....


I prefer my days to be 14-16 hours long and my nights to be 10 hours long

Well, now your days will be eight hours long and your nights will be forty-seven minutes long. It’s about the same thing, isn’t it?

 

[wwjd_kilden]

*skeptical*

 

[resurrection33]

*skeptical*

You’ll learn to love it.

 

[greatkraw]

You’ll learn to love it.

you will learn to love it even if it kills you imagine how much sompler it will be calculating times and dates with a totally metric system you can do things like convert celcius to days

 

[resurrection33]

you will learn to love it even if it kills you imagine how much sompler it will be calculating times and dates with a totally metric system you can do things like convert celcius to days

The human body could be mapped out in cubic meters, and robots could be programmed to do all surgery. You are a programmer—maybe you’re already working on that.

 

[greatkraw]

i once bought a leggo robot kit

never got around to building it

 

[iliveforHim]

YOU'RE ALL GEEKS!!! hmphhh!

MAY I JOIN YOU????!!!! **smiles sweetly**

 

[iliveforHim]

I WANNA BE A KLINGON GEEK TOOOOOOOOOO!!!!!!! AUGHHHHHHHHHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!

**throws self on floor and lands on tummy, kicks feet and punches carpet with fists...**

 

[iliveforHim]

Fine! It's my bed time anyway!!

(I'll be back tomorrow...heeee heeeee)

 

[greatkraw]

Fine! It's my bed time anyway!!

(I'll be back tomorrow...heeee heeeee)

behaves just like a cardassion spy

oops, we havent discovered them yet have we?

 

[greatkraw]

Metric'DAY''''months''''days'''''hours''''minutes'''''''seconds
'1,000.000'10,957.50''365.25''8,766.00'525,960.00'31,557,600.00
'''100.000''1,095.75'''36.53''''876.60''52,596.00''3,155,760.00
''''10.000''''109.58''''3.65'''''87.66'''5,259.60''''315,576.00
'''''1.000'''''10.96''''0.37''''''8.77'''''525.96''''''31,557.60
'''''0.100''''''1.10''''0.04''''''0.88''''''52.60''''''3,155.76
'''''0.010''''''0.11''''0.00''''''0.09'''''''5.26''''''''315.58
'''''0.001''''''0.01''''0.00''''''0.01'''''''0.53'''''''''31.56

so 1000 metric days in a year

8.77 OLD hours in a metric day

actually, with the increased rotation rate, some people will experience weight loss

thos near the equator or in texas will notice they are lighter

people in norway will notice hardly any difference

given a rotation period of 24 hours (and a diameter of 8000 km) becomes a rotation period of 8.77 hours


it should be possible to calculate the increased acceleration at the equator and subtract that from the acceleraton due to gravity 10m/sec

how do we do the sums someone?

 

[iliveforHim]

actually, with the increased rotation rate, some people will experience weight loss

thos near the equator or in texas will notice they are lighter

people in norway will notice hardly any difference

given a rotation period of 24 hours (and a diameter of 8000 km) becomes a rotation period of 8.77 hours


it should be possible to calculate the increased acceleration at the equator and subtract that from the acceleraton due to gravity 10m/sec

how do we do the sums someone?

HU????

 

[iliveforHim]

behaves just like a cardassion spy

oops, we havent discovered them yet have we?

Are cadassion spys really cool, sophisticated, cuddly, smart spys?

 

[Belgian_Pilot]

My weight = 65kg (let's suppose it's my mass m). Actually weight is not mass. Weight is force (my mass will be the same, but my weight will change). Weight is a force F(w)

To calculate v, we suppose I'm at the equator: r = 6.378.000 m (diameter is not 8000km Kraw)
With 1 period = 8h = 28800 sec, v = Pi*2*r/28800sec = 40074249 m/28800sec = 1391 m/sec

So F centrifugal = 65 * 1391² / 6378000 = 19 Newton (direction is off cource away from earth).

Do the same with the 'normal values', you get the current F centrifugal = 2 Newton. Thats a difference of 17 Newton.

So my "weight" would be reduced by 17 Newton

Here you go

 

[iliveforHim]

hmmmmph...show off....I knew that...I was just pretending to be clueless....

(j/k Aksel!!! j/k!!!! )

 

[iliveforHim]

Ohhhhh, I get it, so what you're all saying is, that with the centrivucal intonation of the rammification of the equator, twelve times two equals buckle my shoe, and assuming that the earth spins fivethousand and thirtythree milliseconds per concordination of distansation, the pixels involved are no more than those on the megasypnosis of the arterosclerosis...ahhhhhhh, I see...yes...it's all very clear to me now...

so can I be a Klingon?

 

[greatkraw]

My weight = 65kg (let's suppose it's my mass m). Actually weight is not mass. Weight is force (my mass will be the same, but my weight will change). Weight is a force F(w)

To calculate v, we suppose I'm at the equator: r = 6.378.000 m (diameter is not 8000km Kraw)
With 1 period = 8h = 28800 sec, v = Pi*2*r/28800sec = 40074249 m/28800sec = 1391 m/sec

So F centrifugal = 65 * 1391² / 6378000 = 19 Newton (direction is off cource away from earth).

Do the same with the 'normal values', you get the current F centrifugal = 2 Newton. Thats a difference of 17 Newton.

So my "weight" would be reduced by 17 Newton

Here you go

actually I misled you


earths circumference is 24000 miles = 40000 kilometres
diameter is 8000 miles = 12000 kilometres

radius = 6000 kilometres

or as BP says 6,378,000 metres

so rotational velocity = 40000km/8.77 (old)hours

so we get 40000*1000 metres/60*60*8.77 sec = 1267m/s




previously it was 40,000,000 metres/ (3600 * 24)sec = 463m/s



now we have our 2 velocities

(thank you aksel)

f1 = (m * 1267 * 1267) / 6,378,000 = 0.251*Mass metres/second/second


f2 = (m * 463 * 463) / 6,378,000 = 0.033Mass metres/second/second


assume a mass of 100kg

f1=25.1 newtons
f2 = 3.3 newtons

given gravity is 9.8 newtons INCLUDING the 3.3 newtons

we get new gravity = 9.8 newtons +3.3 newtons -25.1newtons = -12 newtons

therefore we will fly off the planet at the equator at a velocity of 12 metres per second

this will not happen at the north pole or in Norway