Discussion in 'Science & Technology' started by The Imp, May 4, 2010.
Once all your friends and family died, wouldn't you begin to envy the dead?
Link to video.
I think I remember someone posting somewhere a range of spectral types which have the worst flares...
Well I would want to keep them young too ideally.
I'll admit I'm biased, as I find the field rather boring, but I think when people turn to genetics to combat aging, they're missing the forest for the trees. Treatments will come from that approach, but I'm more interested in why it happens in the first place, and questioning if it's a wise thing to meddle with. I surely don't wish to die until I'm ready, and I wish to have a sharp mind and body until that moment, but that doesn't mean I want any of you to live the same life span if it means competing with me for resources.
Knowledge is always primitive if you take a long view on your goals. There's no reason to think you'll be able to stay forever 30, but there's plenty of hope that you'll be able to slow aging down so that you might be have the body and mind of say, someone in their 50s (or what would be a healthy 50s in today's world) well towards the end of your life. All you can do is prepare for it, keep in the best shape you can. If it was in vain, you'll still be in better shape than most of your peers. My grandparents are in their 80s now and can still walk a 5k without draining themselves.
Yeah, that's my plan. Frankly I hope I can be healthier at 50 than I am now. I am cautiously optimistic about my ability to fix my sleep problems, lower my stress levels, regulate my eating & hopefully stay in one place long enough to create a solid social network (finding a career I love would help too).
Regarding Humans on the Moon:
Our bones, muscles, tendons, and all the rest have evolved under the 'g' regimen of Earth. That means, when a human body is subjected to a lesser 'g' value (as on Mars or the Moon), there are certain bio-mechanical effects.
What I'm wondering about is what happens to the legs, ankles, and feet when one jumps. Here on Earth, the body pushes off with a certain force, and the result is that when you land the legs and all are subjected to stresses. But we've evolved to be able to withstand those stresses - which is why you don't get hurt by jumping up and down, but you will get hurt if you jump off a 6 floor building.
But on the Moon 'g' is less. So does this mean that the bones, ligaments, tendons, cartilage, etc. can safely withstand a base jump from a higher altitude? Likewise, if one does a standing jump on the moon, does the landing result in less forces on impact because of lower 'g' - or is it the same force, since the apex of the jump turns out to be higher off the surface?
I have a sneaking suspicion that I've already answered my questions, but haven't thought it all the way through yet... at the same time, these sorts of things are really easy to get wrong!
Being on the moon doesn't change your mass, so for jumping on the spot, you'll generate the same force, you'll generate the same take off speed, you'll just go higher because you decelerate slower once you launch. But you'll still land at roughly the same speed as you take off with, have to absorb roughly the same force as jumping on earth. The only issue is probably with how you react, judging the timing of your landing.
Similar deal for jumping from a height. Same landing speed = same forces to deal with, it's just that on the moon you need to jump from higher to have the same landing speed. At least for humans at low speed, anyway. For an animal that can actually survive landing at terminal velocity on earth, whether it's an insect or a human with a parachute, they're going to reach a higher speed on the moon thanks to not having the air resistance. Jump off a 3km high cliff on earth, and you'll either land at ~200 kph without a parachute or fairly slowly with one. Go higher and that won't change. Jump off a 3km cliff on the moon, you'll land at ~360 kph regardless, and going higher will mean landing at a higher speed.
As in space, living on the Moon would weaken us, because it takes less force to hold our bodies up, and do simple things like walk. sanabas is right that if you push off with the same amount of force, you land with the same amount of force, but on the moon you would not want to push off so hard. Likely an astronant living a while on the Moon would get used to maving by pushing off with a lot less force, so that he does not moonjump like in the Apollo mission.
So as in space, additional excersize would be needed to maintain our bodies on the Moon.
Anyone know how to formulate a rebuttal to this assertion? I think it involves frames of reference, but I can't think of anything else.
You add the velocity of the vehicle you are travelling on to your own. It's only if you jump while the reference frame is accelerating/decelerating that your position changes.
It's safe to jump on the back of a truck, you'd just fall straight down, so the second paragraph is bollocks. Same with throwing a ball on a train.
EDIT: It's just Newtonian mechanics,
v = u + at
Thanks. It's been too long since I've had a physics course.
Don't try jumping on the back of a truck if it is windy (no friction from the truck bed) or if it is accelerating or decelerating though
Sage advice indeed.
Yeah, the essential difference in the two examples is that in the first, the air that you're jumping through is also rotating at (whatever) speed, while in the second example the flatbed (and you) are moving faster than the air - so you slow down as soon as you leave the bed of the truck.
If the truck were driving downwind at 30mph in a 30mph wind (perceived windspeed would be 0mph), then you would indeed land precisely where you were.
The normal way this is explained is that you can juggle on a train in exactly the same way as you can juggle while "standing still".
Ack! You nerd-sniped me out of blissful lurkerdom.
I didn't remember until I just did it now how simple that proof is though:
Let S = 1+x+x^2+...
Then S = 1+x+x^2+x^3+... = 1+x*(1+x+x^2+...) = 1+xS
Solve for S --> S = 1/(1-x)
Let x = -1.
Then S = 1 - 1 + 1 - 1 + ... = 1 + -1( 1 -1 + 1 -...) = 1 - S
S = 1/(1-(-1)) = 1/2
You can only do the algebra on the formal power series if it converges absolutely, i.e. if it converges with x replaced by |x|.
That means it only works if -1 < x < 1
Yeah, I knew that, but I forgot to include that disclaimer in there. Obviously you can "prove" all sorts of nonsense if you let |x| >= 1.
I wasn't doing it for a converging, infinite sum though. I was deriving the formula for a + ar + ar^2 + ... + ar^n
Then the infinite, converging series is just a simplified version of that.
Well just use Wolfram Alpha
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