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Just his usual overacting
GCS d--(d-) s-/++ a C++++ U+++ P- L+@ E-- W++ N+ o+ K- w+++ O? M-- V? PS+ PE- Y+ PGP t+ 5? X R+++ tv-- b+(+++) DI+++ D++ G e++ h--- r+++ y+++* Weapons extension: ma- k++ F+2 X
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If that doesn't kill him, nothing will.
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He went up and came down again - safely! Yay!
- I would love to change the world, but they won’t give me the source code.
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He survived!  I was worried about the g forces he would encounter during launch, given his advanced age.
EDIT:
But then again, since they didn't go into orbit, the rocket probably accelerated only to a few hundred miles per hour, so the g forces would not have been that bad.
The difficult we do right away...
...the impossible takes slightly longer.
modified 13-Oct-21 11:44am.
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Pretty sure I saw 2230mph (3590 km/h) just before the booster shut off.
Also, pretty sure I heard somewhere that the meat bags in the capsule would see about 4g's - but can't find anything online to prove it.
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As Spock would say; "It is logical"
The less you need, the more you have.
Even a blind squirrel gets a nut...occasionally.
JaxCoder.com
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I was just sorry he didn't wear one of his Starfleet uniforms for the flight.
Money makes the world go round ... but documentation moves the money.
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I teach high school maths to a child, and yesterday, there was some spare time as it was raining outside, and he could not head back home. So, I opened my old book of math puzzles and gave him some puzzles. One puzzle is quite interesting, and is said to have occupied an acquaintance over a two-and-a-half hour train journey from Mysore to Bangalore, two cities in Southern part of India separated by about 150 km. The puzzle goes like this:
Four Fours
Express any number between 1 and 100 in terms of four 4s, and mathematical symbols, + - x / . .4-dot ! sqrt, etc.
For example,
1 = (4 + 4)/(4 + 4)
4 = 4 / sqrt(4) + 4 / sqrt(4)
12 = (4!/4) + (4!/4)
36 = (4!/4) x (4!/4)
45 = 44 + 4/4
100 = (4/.4) x (4/.4)
Can you attempt to fill in the other numbers from 1 to 100. Some numbers have non-unique solutions.
One challenge here is to find HTML symbols to express the numbers. For example, .444444... is expressed as .4-dot (with the dot on top) - so this fraction 4/9 is expressed as .4-dot, with only one 4 being used. Also, expressing square root needs a unicode symbol in HTML.
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I did this back in 1482 when I was at skewl.
We just got + - x / ! and sqrt.
4!/4 featured heavily, 4/.4 is just cheating
veni bibi saltavi
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imho, there is nothing of educational value in this puzzle.
By the way, i've been on that train (1976); scenery's too nice to miss.
«The mind is not a vessel to be filled but a fire to be kindled» Plutarch
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Did that route in a hired car in 2003. My VP of Engineering (a Sri Lankan Tamil) wanted to visit Chamundi's temple, so we made a day trip of it and also visited the Mysore Palace. Both impressive.
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Kind of unusual to hear a Canadian has done that trip, and that too by road!
I've never done it by road but I've taken the train, just once.
Cheers,
Vikram.
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I also did the round trip by road a few years ago, but in a taxi. I would never drive myself on those roads.
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More of recreational value, something like a mathematical excursion.
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A variant using the four digits, 1 9 7 and 2, in order, was educational to me. Took a few minutes to write a BASIC program to generate and evaluate random Polish notation expressions. The program took days to generate almost all of 1..100 on a time-shared mini-computer. When one of my kids had a similar assignment in school, I wrote the program again in C for modern hardware. Ran instantaneously.
That program taught me that computers can solve problems without human super-skills being needed.
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Amarnath S wrote: any number between 1 and 100
Presumably that means any natural number? If you had to express every real number, or even all the rational numbers, in that range, you'd need a much longer train journey! 
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Yes. I meant natural numbers. Not rational numbers, that set may be countable but is infinite, or irrational numbers; that set is not even countable.
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16=4+4+4+4
88 = 44+44
2=4/4+4/4
4=4+4*0+4*0+4*0
...
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Last one - cannot use the number 0 in the expression.
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Ok
32=4*4+4*4
36=4!+4+4+4
56=4!+4!+4+4
76=4!+4!+4!+4
96=4!+4!+4!+4!
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Since this is "codeproject" my train trip was short. Took about 15 minutes to get the 1st order solution in C.
Yes...I know it's simplistic....but a good exercise nonetheless. Obviously could be improved for all the special cases....could obviously compute squares, square roots, factorials, etc and start from the largest of those to reduce the value. I'll leave that exercise for others.
In this code just change the modulus value to whatever you want and you'll get all the 1st order solutions.
#include <stdio.h>
int modulus = 5;
void do4(int n)
{
int mod = n % modulus;
int i, j;
printf("%d=", n);
for (i = modulus; i <= n; i += modulus)
{
if (i > modulus) { printf("+"); }
printf("%d", modulus);
}
for (j = 0; j < mod; ++j)
{
if (i != modulus) { printf("+"); }
printf("(%d/%d)", modulus, modulus);
}
printf("\n");
}
void main(void)
{
int i;
for (i = 1; i <= 100; ++i)
{
do4(i);
}
}
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Four fours! (not arbitrarily many)
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It works for four fours. But this is "Code Project" and writing a routine to do just that one example is no fun.
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The output, after repairing a small bug - condition in 2nd loop is i != modulus || j > 0:
1=(5/5)
2=(5/5)+(5/5)
3=(5/5)+(5/5)+(5/5)
4=(5/5)+(5/5)+(5/5)+(5/5)
5=5
6=5+(5/5) etc.
None of these lines has five fives, so this does not solve the problem. Running it with modulus 4, one gets:
1=(4/4)
2=(4/4)+(4/4)
3=(4/4)+(4/4)+(4/4)
4=4
5=4+(4/4) etc
Here, only the solution for 2 has four fours, all others are not solutions (two fours, six fours, one four, three fours, ...). So how does this "work for four fours"?
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