cjohnso0

I sort of agree with doG. Sort of becuase I had it right on the first try, but looking back, doG's observation makes sense.

The current record (from the NHRA) is 4.428 by Tony Schumacher back in '06. This is a Top Fuel dragster I believe. I'm looking for a reliable source for the 3 second rocket cars still.

Neat, It's definately not faster for numbers with larger digits. Try 9999 x 9999, not only do you need a big sheet of paper, it gets confusing real fast. It seems to take about the same amount of time (as standard multiplication) for smaller numbers though.

Assuming it works, what are the benefits of this over the standard crankshaft we all know and love? Along with a working design, this is a pretty important question.

This one's got me stumped so far

river_rat, I'm following you now, I was missing that you were referring to the original problem posted. All of my work was for the .pdf version. After playing with the original question, I cannot find a solution, at least not one that makes sense, yet. by this I mean a mathematical solution for the .pdf question. If the original question posted is correct, there must be some obscure solution that's beyond me at this point.

I think your sketches need some more detail, for example: Where is the combustion chamber? Assuming the crank is rotating CCW, what happens to the piston after 'B', it looks like it will fall back onto the springs, then when the crank comes around, it will tear the teeth off on the left side. I'm also not sure that gear teeth can take that kind of a sudden impact load without immediately shearing off or prematurely failing in fatigue. For a proper analysis, you really need to flesh out your design more.

river_rat, In your example, your system of equations does not have a solution. If you were to start with a solution, you could then build any number of equations which would satisfy it: x,y,z = 1,1,1 x + y = 2 x - z = 0 y + 2z = 3 2x - 5y + z = -2 Given these 4 equations, you can remove any one and still get the same solution, assuming the equations you choose have at least one occurence of each variable. I just ran the numbers using 3 different sets of 23 from the available 25, and the solution came up the same. I'm not going to try all 2300 I'm still not sure that there will be a solution to the initial problem.

My methodology started out poorly... 1st up was to boot up excel, my tried and true numerical analysis program. Then I simply laid out a matrix of all 23 used letters, each row being an equation for one of the names. Then I attempted gaussian elimination. This is when I saw that Y can equate to anything I choose, without effecting the rest of equations. At this point, my answers were way off base, so I tried using the solver tool in excel on the first 23 names (23 becuase I only had 23 potential letters) (got the method from googling "Excel Simultaneous Equations", initial guess was 1 for all letters) Bammo! I got my answers in seconds So I tried the linear algebra, but failed, then used my easy tool to solve. Given the problem definition, I'm still not convinced that there is a definate solution, or that the missing letters are of any consequence. We'll need to see if anyone can find a pattern in this. Not sure if I can, but I can post the excel file if anyone wants, or e-mail, as that will work.

Whoa - Just looked at my data in my last post and the numbers go 2-24, omiiting 19. If Y = 19, then Feynman = 94 thoughts? M 2 T 3 V 4 H 5 A 6 E 7 O 8 I 9 U 10 B 11 G 12 K 13 P 14 S 15 F 16 W 17 Z 18 Y 19 C 20 L 21 N 22 R 23 D 24

Just chiming in here, i've been thinking / working on this one for a week now, and finally i've come to the conclusion that the answer is inderminate. Given the problem specifications, and using the data from the .pdf linked above, the letter Y can have any value. Methodology was to make a huge matrix and reduce until I had some equations I could solve, at that point I saw the problem. Only 1 occurance of the letter Y. Here's values I've gotten for all the others: A 6 B 11 C 20 D 24 E 7 F 16 G 12 H 5 I 9 K 13 L 21 M 2 N 22 O 8 P 14 R 23 S 15 T 3 U 10 V 4 W 17 Z 18 Using these values, all names add up correctly to the .pdf values, and Y can have any value. Thinking about it, I would be not be surprised if the answer is something like this, as they can't expect you to solve 23 equations at once, high IQ or not. Anyhow, that's just my take, hopefully there's not too many errors...

Engineers Edge is a pretty good one, it's got a lot of mechanical engineering related stuff, calculators, graphs, etc... And not totally engineering, but http://www.treasure-troves.com/ has a ton of science and math, I use the math section pretty often as an engineer. As stated above, give a clue on what you need, and I'm sure a few more resources will be provided.