Wikipedia:Reference desk/Archives/Mathematics/2013 September 27

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September 27

How fast can humans accelerate without dying from G-force

Hi refdesk peeps,

In conversation with friends, we wondered just how fast a person could theoretically travel to planet Mars. I did a little bit of research, and learned that it's unlikely anyone could live through more than about 9g for more than a few minutes - that probably needs to be way lower, but for the purposes of my calc, let's say 9g perhaps. But I don't the rate of acceleration that would result in 9g.

We're ignoring fuel requirements and propulsion, and all that stuff. And being super-lenient on survival. I'd just like to know, if you accelerated as fast as possible for 'a while', and then decelerated, how long would it take to travel the roughly 200 million km to the red planet.

Would the resulting high-speed actually be a problem, or is it just G that'd be the killer?

This isn't homework - it's just a silly pub-conversation. Thanks in anticipation of any replies, — Preceding unsigned comment added by 88.104.6.199 (talk) 04:03, 27 September 2013 (UTC)

For a pub, a back-of-the beer coaster estimate is appropriate. Distance in terms of acceleration and deceleration times would be $d=at_{a}^{2}/2+at_{d}^{2}/2$. If we set $t_{a}=t_{d}$, then $t_{a}={\sqrt {d/a}}$ for total time $2t_{a}$. Putting in the numbers (in meters) $d=2\times 10^{11},a=10g$ where $g$ is the number of gees, the total time is $t=2.8\times 10^{5}/{\sqrt {g}}$. Of course the highest velocity would be $v=at_{a}=1.4g\times 10^{6}m/sec$, or a half of a percent of the speed of light at one gee. It would be quite a trip :-) --Mark viking (talk) 04:32, 27 September 2013 (UTC)

Thanks very much, that's great :-) So if I assumed we could somehow survive 9g for a bit, I make it 2.8*10^5 / 3 = 93,333s - around 26 hours. I hope I got that last bit right. At a more 'realistic' (for want of a better word) 3g, about 45 hours. (Kinda surprisingly not massively different). Now we just need incredibly OP weightless propulsion, a ship that can handle it. Cool, thanks again. — Preceding unsigned comment added by 88.104.6.199 (talk) 06:08, 27 September 2013 (UTC)

At these speeds, particle shielding would become a problem. Solar wind is mostly a particle stream at 0.002c , and at 1% c, one would go through five times the number of particles per second than at 0% c. These particles would have more kinetic energy, too. But I digress; it should probably be asked at RD/S if somebody wants to know more about that. - ¡Ouch! (^{hurt me} / _{more pain}) 12:34, 27 September 2013 (UTC)

Note that how many g's you can take very much depends on the duration. At 9g, for example, your blood probably won't circulate, so you have only seconds or minutes before you pass out, then die. A flight suit that contracts and relaxes to help pump your blood might help somewhat, but we're still only talking minutes, not hours. As for what acceleration a human could withstand for hours, we might be down to 1.1 or 1.2 g's, to avoid permanent damage, similar to bed sores. StuRat (talk) 16:30, 27 September 2013 (UTC)

Also note that the problem isn't caused by the acceleration rather that the body is being accelerated by contact forces. The contact force ends up being transferred into your body due to the body deforming and stresses building up in your body as a result. If you could apply a force uniformly to your body, you would not feel the acceleration. One example is gravity, you are accelerated at 100 g due to a uniform gravitational field, you will feel weightless.

Top accelerate a person safely at 100 g's, you could consider placing a large number of small magnetic pallets in the body of the person. The person could then be let to float in a strong magnetic field in the accelerating spacecraft. You could even try to regulate this system so that the magnetic field contributes to one g less than the spacecraft's acceleration. The contact forces will then yield the remaining 1 g, making the person feel like he's on Earth. Count Iblis (talk) 17:35, 27 September 2013 (UTC)

Liquid immersion would increase the tolerable acceleration to around 15 to 20 g's. Above that one needs a breathable liquid with suitable density (close to the density of water). Ssscienccce (talk) 17:06, 28 September 2013 (UTC)

Viability of a certain hash function?

Suppose we have two large prime numbers (say 128 bits each) "BASE" and "MODULUS" which are publicly known, and then some secret number of arbitrary length "EXPONENT". How secure is the result ((BASE ^ EXPONENT) % MODULUS)? 70.112.97.77 (talk) 09:46, 27 September 2013 (UTC)

I would think not very secure, because of the following:

With the base and the modulus you can compute ord_modulus(base), and then finding the exponent only requires checking multiples of ord_modulus(base), which I believe could be done relatively quickly. -- Toshio Yamaguchi 12:39, 27 September 2013 (UTC)

Okay, thanks! Your explanation eventually led me to the page on Diffie-Hellman (a very similar problem, apparently), which points out:

"The protocol is considered secure against eavesdroppers if G and g are chosen properly.[...]The order of G should be prime or have a large prime factor to prevent use of the Pohlig–Hellman algorithm to obtain a or b. [...] g is then sometimes chosen to generate the order q subgroup of G, rather than G, so that the Legendre symbol of g^a never reveals the low order bit of a."

So the above implies that, considering the group "G" of _MODULUS(BASE), as long as I can (1) verify that there is some integer "E" such that BASE^E ≡ 1 (mod MODULUS) (thus proving that the order of G is prime) and then (2) find (and replace BASE with) the generator "D" of a subgroup of G (in order to protect the lower order bits of EXPONENT), then my algorithm (ie: D ^ EXPONENT % MODULUS) should be secure as well. Does that sound correct? (Sorry, my number theoretic abilities are somewhat lacking - I'm more of a programmer than mathematician.) Solving requirement #1 seems pretty straightforward, but I'm a little confused how to go about solving for #2 (or what it even means, really!). 70.112.97.77 (talk) 15:39, 27 September 2013 (UTC)

I think what you are describing is related to the discrete logarithm problem. If this is correct then your method is moderately secure, as there is no known algorithm for solving the problem in a time that is a polynomial function of the number of digits in the modulus. However, there are algorithms that can solve the problem in a time that is proportional to the square root of the modulus, so a 128 bit modulus is not completely unbreakable - see discrete logarithm records. A 256 bit modulus would be better. Gandalf61 (talk) 16:01, 27 September 2013 (UTC)

Interesting, but then the article you linked suggests that even a 1024 bit modulus wouldn't necessarily be sufficient. I was hoping for quite a bit more security, something requiring, say, the lifetime of the universe using all of the computing power in the world to break! Any idea how many bits would be required to achieve that? 70.112.97.77 (talk) 16:40, 27 September 2013 (UTC)

That would depend very much on what you mean by all the computing power in the world: the computing power that exists today, or the power that one would expect in the future based on Moore's law. Assume that processing power doubles every 18 months, and we have a key length that would take 10¹⁴ years to break (the estimated lifetime of the Stelliferous Era) with todays computing power: 10¹⁴=2^46.5 and 46.5*1.5= 69.75, so in 70 years time we will have enough computing power to break that same encryption in less than a year. Ssscienccce (talk) 17:33, 28 September 2013 (UTC)

Good point. Well that settles it, I guess I just need to use googolplex-bit numbers! Time to optimize that Miller-Rabin code... 70.112.97.77 (talk) 01:11, 29 September 2013 (UTC)

See also RSA (algorithm). Duoduoduo (talk) 16:31, 27 September 2013 (UTC)

Determining a coefficient

I know that the coefficient of x^k in (1 + x + x^2 + ... + x^k)^n is n+k-1 C k. Is there a similarly easy expression for the coefficient of x^k in (1 + x + x^2 + ... + x^r)^n, where r<k and nr>=k? 31.54.112.70 (talk) 11:51, 27 September 2013 (UTC)

It is:

$\sum _{p=0}^{\frac {k}{r+1}}(-1)^{p}{\binom {k+n-1-p(r+1)}{k-p(r+1)}}{\binom {n}{p}}$

Count Iblis (talk) 14:09, 27 September 2013 (UTC)

Thanks. Is the upper limit of the summation the largest integer not exceeding k/(r+1)? 31.54.112.70 (talk) 15:13, 27 September 2013 (UTC)

Yes, you can also say that the summation stops when p exceeds k/(r+1). Count Iblis (talk) 15:27, 27 September 2013 (UTC)

I believe the floor notation could be used to state it explicitly. $\left\lfloor {\frac {k}{r+1}}\right\rfloor$ lulzmango (talk) 07:58, 28 September 2013 (UTC)