#### Re: "Light signals can sometimes take up to 20 minutes to journey from Earth to Mars "

Perhaps it's referring to when Earth and Mars are at their furthest apart along their orbits vs their nearest. (or points in between)

14 posts • joined 7 Dec 2013

Wednesday 18th December 2019 18:28 GMT #### It takes time for A to affect B, even if you can't measure it.

Assuming A and B are different, It takes time for A to affect B, even if A's affect on B is quicker than anything you can measure.

If A affects B (aka B is affected by A) sooner than (distance between A and B) * (1/'the speed of light'), something faster than 'the speed of light' has occurred.

Thursday 6th December 2018 03:41 GMT #### "between 10 and 30 times lower than CMOS", math???

"and consequently lower power (between 10 and 30 times lower than CMOS)."

10 times lower than a number, say X is X-10*X = -9*X.

Notice the sign, the before has opposite sign of the new. Now are we now gaining power by using this, or was it gaining power before and now it's losing power, or was it zero power change the whole time? or perhaps someone(me?/you?) doesn't understand how this math/power usage works? This doesn't make since to me. What is it really?

Does the same apply to

"low voltage (as much as five times below today's CMOS-based chips)"

5*below a number, say X, is X-5*X =-4*X

Thursday 6th December 2018 03:31 GMT #### "between 10 and 30 times lower than CMOS", math???

"and consequently lower power (between 10 and 30 times lower than CMOS)."

10 times lower than a number, say X is X-10*X = -9*X. Notice the sign, the before has opposite sign of the new. Now are we now gaining power by using this, or was it gaining power before and now it's losing power, or was it zero power change the whole time? or perhaps someone(me?/you?) doesn't understand how this math/power usage works? This doesn't make since to me. What is it really?

Monday 9th December 2013 19:07 GMT #### Re: and now for some math...

I suppose that could count as a star too... I should have said excluding the Earth's sun, or something like that...

The Earth's sun is ≈ 1/278896.8645 = 1/((4.2421 ⋅ lightyear) ∕ (8 ⋅ lightday ⋅ (lightyear ∕ (365.25 ⋅ 24 ⋅ 60 ⋅ lightday)))) times as far as the next closest star from Earth.

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Counting this sun as a star, and 1 AU as the distance between the Earth and it's star, the sun.

"Astronomical unit, an approximation for the average distance between the Earth and the Sun"

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

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// Qalculate

AU⋅13242⋅day/(127.03⋅AU) = (astronomical_unit ⋅ 13242 ⋅ day) ∕ (127.03 ⋅ astronomical_unit)

≈ 9.00660316460679 Ms

≈ 9006603.16460678579863 s

≈ 0.285402031986171 year

---

// TI Voyage 200

_au*13242*_day/(127.03*au)

≈ 9006603.1646068*_s

≈ .28540812789648*_yr

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Qalculate, TI Voyage 200, and Wikipedia, use different values for AU

149,578,706,600 m, Qalculate v.0.9.7

149,597,900,000 m, TI Voyage 200 OS Version 3.10, 07/18/2005

149,597,870,700 m, http://en.wikipedia.org/wiki/Astronomical_unit (the same day as this post)

(looks like the TI Voyage 200's AU is a rounded version of Wikipedia's AU, and Qalculate's AU uses a different measurement)

Sunday 8th December 2013 09:48 GMT #### and now for some math...

(If something is wrong with this, please correct me)

---

Given:

127.03 AU approximate distance traveled so far by Voyager 1

4.2421 ly is the approximate distance to nearest star from Earth

Find:

How long would it take at Voyager 1's average speed so far, to go the distance of Earth to Earth's nearest star?

--- ---

My Answer:

Approximately 76568 years. Last digits maybe off due to rounding, etc...?

Although, I got 76576 years from Qalculate with units, and 76566 years from Ti Voyage 200 with units.

--- ---

// Research:

127.03 AU is approximately 0.7336640935983627 light days

127.03 AU is approximately 0.002008662816152567 light years

From September 5, 1977 to December 7, 2013 is 13,242 days

---

// Using Qalculate without units:

(4.2421ly)*(13242d)/(.73366ld*(ly/(365.25ld)))=___

(4.2421 ⋅ 13242) ∕ (0.73366 ∕ 365.25)

≈ 27965968.79351470709

days

((4.2421 ⋅ 13242) ∕ (0.73366 ∕ 365.25)) ∕ 365.2425

≈ 76568.22191698586

years

---

// Using Qalculate with units:

4.2421⋅ly⋅13242⋅day/(127.03⋅AU) = (4.2421 ⋅ lightyear ⋅ 13242 ⋅ day) ∕ (127.03 ⋅ astronomical_unit)

≈ 2.4165557924 Ts

≈ 76.57603215692 kiloyears

// Qalculate Uses

ly = lightyear ≈ 9.46073047258 Pm

day = day = 86.4 ks

AU = astronomical_unit = 149.5787066 Gm

year = year = 31.5576 Ms

---

// Using Ti Voyage 200 with units:

4.2421*_ltyr*13242*_day/(127.03*_au)

≈ 2.41619E12*_s

≈ 2416194140896.8*_s

≈ 76566.207457403*_yr

// Ti Voyage 200 uses

1*_ltyr = 9.4605284048794E15*_m

1*_day = 86400.*_s

1*_au = 149597900000.*_m

1*_yr = 365.24219878125*_day

--

// Web Sources:

http://en.wikipedia.org/wiki/Voyager_1

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

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

http://en.wikipedia.org/wiki/Julian_year_%28astronomy%29

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

http://www.convertunits.com/from/AU/to/light+day

http://www.convertunits.com/from/AU/to/light+year

http://www.timeanddate.com/date/durationresult.html?m1=9&d1=5&y1=1977&m2=12&d2=7&y2=2013

---

// Calculator Sources:

Voyage 200, Qalculate

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p.s. for those who want more digits in light days for my calculation...

((4.2421 ⋅ 13242) ∕ (0.7336640935983627 ∕ 365.25)) ∕ 365.2425

≈ 76567.79469211468

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