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That girl is either very widely read, or needs either a psychologist or an IQ test!
Yes , maybe she was somewhat older.Tom.G said:That girl is either very widely read, or needs either a psychologist or an IQ test!
If you want to go past the castle, your logistical support may have to go back and forwards past it and be subject to raiding from the castle. If there are significant numbers of troops inside they may raid your army (and/or your supplies) from the rear at the same time as you are engaging somebody in front of you. Or they may simply note that you and your army are no longer between them and your country, which might be bad. And if you want to occupy the land you need to do so in sufficient numbers that your garrisons are all strong enough to make a go at resisting a raid from the castle, or small garrisons will get picked off one by one. In short, the problem isn't really the castle so much as the fact that it's a shelter for enemy armed forces.fresh_42 said:Simply ignore that thing! Why wasn't that ever a possibility?
Ill Bayeux some embroidery.DrGreg said:The Bayeux Tapestry is not a tapestry.
It's an embroidery.

I think we should keep this in mind.The people from ten thousand years ago weren't so much different from us. They have had the same capabilities and mental awareness as us. Biology didn't evolve so much during that relatively short period. All that changed is our technological evolution and our ways of sharing information.
And how long did it take you to calculate ##\log_2 5## by hand?WWGD said:Kind of surprised to see this ( alleged?) problem in the entrance exam for Cambridge Math program: Find ##x## if ##8^x+ 2^x =130## . It took me all of 2 minutes to figure it out. Just use that ##8=2^3##.
I doubt you're required to. A Taylor series would likely give you a good approximation. Since ##2^2=4; 2^3=8## , it is closer to ##2## than to ##5##. Use the derivative for a tangent line approximation.fresh_42 said:And how long did it take you to calculate ##\log_2 5## by hand?
Power series can be nasty. You need half a googol steps to calculate ##\pi=4\tan^{-1}(1)## up to ##100## digits from the standard power series, and ##90## with a simple trick.WWGD said:I doubt you're required to. A Taylor series would likely give you a good approximation. Since ##2^2=4; 2^3=8## , it is closer to ##2## than to ##5##. Use the derivative for a tangent line approximation.
Not to compute the Taylor series for ##Log_2 x## about ##5##. Are you sure they require anything beyond Edit:##ln(5)/ln(2)##?fresh_42 said:Power series can be nasty. You need half a googol steps to calculate ##\pi=4\tan^{-1}(1)## up to ##100## digits from the standard power series, and ##90## with a simple trick.
I was joking.WWGD said:Not to compute the Taylor series for ##Log_2 x## about ##5##. Are you sure they require anything beyond Edit:##ln(5)/ln(2)##?
I think you're off by two orders of magnitude and a unit change. I make it ##365×24×3600×3×10^8\mathrm{m}## is 9.4608 trillion kilometres, or about 5.913 trillion miles.WWGD said:Ok, not too hard to remember: a light year is 946 trillion milles , plus 80 million miles, i.e., 946.080.000.000
Ah, I guess I thought the speed of light was given in miles/second. I'm remembering now it is 186,000 miles/sec, not 300,000 miles/sec.Ibix said:I think you're off by two orders of magnitude and a unit change. I make it ##365×24×3600×3×10^8\mathrm{m}## is 9.4608 trillion kilometres, or about 5.913 trillion miles.
As a " meta point" of sorts, these casual exchanges can be valuable. This one allowed me to correct my knowledge base.WWGD said:Ah, I guess I thought the speed of light was given in miles/second. I'm remembering now it is 186,000 miles/sec, not 300,000 miles/sec.
So Ill settle for 946 trillion km + 80 million Km. Then I use that 1 mile~1,609 metres. Or metres jn UK ish.
That's 100ly, though.WWGD said:So Ill settle for 946 trillion km + 80 million Km.
If we differed by three orders of magnitude, I'd agree. I think pretty much everybody uses the American standard now anyway. Despite it making less sense.fresh_42 said:I get dizzy. How can you talk about trillions when you are from countries that define it differently?
Oh dear...WWGD said:This source cites it as ##9.46 \times 10^{12}##km. I saw it computed in a YT short and dIdnt double-check.
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You mean ##6## as in ##10^{12}## (US) versus ##10^{18}##?Ibix said:If we differed by three orders of magnitude, I'd agree.
Definitely not here, and I cannot see that anyone quit using -iards.Ibix said:I think pretty much everybody uses the American standard now anyway.
Well, any power of ##10^3## difference rings "maybe it's a mega/giga/tera mix up" alarm bells for me, but yes you're correct here.fresh_42 said:You mean ##6## as in ##10^{12}## (US) versus ##10^{18}##?
I think the battle in English was lost long ago. The only "normal" usage of such values is financial, and that seemed to settle on the US standard when I was a kid.fresh_42 said:Sorry, I thought the British would have used the European standard.