Re: please America .....
The answer to this question can be either "literally" or "figuratively" take your pick.
20 publicly visible posts • joined 7 Dec 2013
Samsung’s spiel also reveals that the display used by Lenovo grows from a 5:4 aspect ratio to an 8:9 ratio – but without disclosing resolution.
numbers to help compare
45:36 aspect ratio to an 32:36
40:32 aspect ratio to an 40:45
{ 2000 x 1600 to 2000 x 2350 } /50 = 40 x 32 to 40 x 47
Was some rounding done on ratios given? 47 != 45
Ok, so maybe the 'blink' idea isn't remotely feasible.
Even so, some misc related thoughts:
It could be preemptive. Like hey it looks like there could be something coming, erroring on the side of caution blink.
Did you know you can still see with your eyes shut? When your eyes are shut you can still tell when the lights are on or off.
Lazers
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There's still the 'shell' during set up idea.
"And the system must run from solar panels, must run cold to not disturb the telescope, must emit no vibration and have no real moving parts (momentum, angular momentum) and can not emit any kind of exhaust or anything which might land on the mirror or optics."
This doesn't have to be attached to the telescope. It could be like one of those plastic Easter eggs shells that holds a toy inside that's not attached to it. Throw away the shell when the toy is ready to be taken out.
Or perhaps it could turn into something like a turtle shell, giving the telescope a place to hide for protection if something major was spotted coming it's way. Although, would moving out of the way be better? How often does something like this occur?
How about for added protection during setup, something like some sort of removable skin/shell/bubble/eyelid that will go into a protective position while the telescope is setting up, then get out of the way when the telescope is ready. And also be able to get quickly get back into/out of protective position if danger is detected/gone, like a blink.
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.
"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
"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?
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)
(If something is wrong with this, please correct me)
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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?
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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.
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// 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