The denominator then needs to be rationalized by multiplying top and bottom by the reciprocal of 1/2 (2). The answer is then equal to 4/2= 2!! It truly does take twice the person.

Ok, I checked with my girlfriend who is working on her phd in algebraic topology. She lost me pretty quick, but there is a branch of math known as abstract algebra. In abstract algebra there is no division or subtraction. Instead you consider the inverse of multiplication and subtraction. So, assuming that you are using the field R, and you wish to divide 2 by 1/2, you must actually multiply 2 by the inverse of 1/2. In abstract algebra, a number and its inverse must multiply to unity. So the inverse of 1/2 is 2. Therefore, we have 2x2=4. Of course you could define a new number system, where what you have said is true, but it is paramount that the new system be consistent. She mentioned the requirements to be consistent to me, but explained that depending on what you are trying to do, the requirements may be looser than those she told me. I assumed that you were using the field R (real numbers) anyway. Hope this helps.
-g
how many times the man does it take to mow the lawn?

Re: Math proves it takes twice the man to ride half the bike

— JonnyD <JonnyD.3lbsm@timelimit.unicyclist.com>
wrote:
>
> Harper’s saying can be represented like this:
>
> 2 (person)
> – (over)
> 1/2 (half the bike)
>
> or
>
> 2
> –
> 1
> –
> 2
>
> (fraction)
>
> The denominator then needs to be rationalized by
> multiplying top and
> bottom by the reciprocal of 1/2 (2). The answer is
> then equal to 4/2=
> 2!! It truly does take twice the person.
>
>
> Cheers,
>
>
> –
> JonnyD - Not weird: Gifted
>
> -Jon Davis-

Dude, I think you screwed up your math.

2/(1/2) = (2/1)/(1/2) = (2/1) * (2/1) = 4

i know that thats right up to there, but i dont really
see what its proving. Although i like where this is
going…

1/(1/2) = (1/1)/(1/2) = (1/1) * (2/1) = 2

by that math, 1 person, over 1/2 the bike = 2!

so therefore one person riding half the bike, takes
twice as much skill: it takes twice the man to ride
half the bike!

let me guess, u and your girlfriend dont get to talk much

once again thanx for a very entertaining thread
in this case also serving to remind me why i dropped maths when i did
interesting tho, i sometimes see similar ‘pictures’ while juggling that i used to c while trying to solve quad equations (i think that’s what they were)
and when i try n explain to someone that u do have points of stability while riding a uni, three of them in fact
the seat and the two pedals
creating a triangle
our most stable shape (next to a circle but we aren’t into monocycles, are we?)
the only ‘problem’ is that, with the cycling motion, the sides of the triangle are never constant and to maintain some semblance of stability, the rider must constantly adjust the amount of power applied to each of the two sides
by this time they normally wander off to go mow the lawn…

I would wonder if it can or should be extrapolated from “it takes twice the man to ride half the bike” to mean that two men riding a tandem bike have in total twice the manliness as one man on a unicycle.

Re: Math proves it takes twice the man to ride half the bike

Reply-To: “Danny Colyer” <danny@jugglersafety.net>
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willfcc wrote:
> Real unicyclists mow the lawn while drinking beer on their uni.

I really wish you hadn’t suggested that. My lawn needs mowing and I’m not
sure I can resist the challenge.

–
Danny Colyer (remove safety to reply) ( http://www.juggler.net/danny )
B4/5v c(+) rv d m(+) w++ q+ k e+ t+ (s) g+ f - http://www.lpbk.net/jc/
“I don’t think proofreading is adequate. All posts should be waxed and
buffed. Then they should wear little tuxedos.” - Greg Harper on usenet

that’s really nice so I hate to say anything about it but…
work = force times distance. acceleration
There are two forces acting between the rear wheel and the ground. The normal force, which does no work, and the frictional force that is caused by the torque applied by the rider. this is the same case for the unicycle. The front wheel, unless braking, has no frictional force acting (assuming planar motion) The front wheel does no work under acceleration. The limiting case is that of all the weight of the rider plus the bike plus some rotational effect all acting as a normal force on the tire. and (mu)s*Fnormal being maximum frictional force attainable. In the case of the uni, this force traction force is destined to be bound to a lower value (if the uni weighs less than a bicycle plus the rotational affect). deceleration
Under braking the front does much work. The maximum work it can do is again limited by the normal force on the ground. Which is again, rider+ride+rotational affect.
It seems the bike has the potential to do more work. But does it? The minor corrections to the unicycle all take work. A unicycle standing perfectly still without falling over, is in fact a valid solution to the equations of motion. It is of course unstable. These corrections to keep a unicycle rolling may stack up to greater than or less than the force of bicycle riding, it just depends. I hope I got this all right.
-gauss
sorry, I just can’t help it.

This thread proves that unicyclists aren’t all scary clowns!

A bunch of them are pretty smart (smart enough that I get totally lost while trying to read, sort of like listening to my english teacher talk about the life's of romantic poets).

I can juggle and drink on a unicycle although not beer cos it’s gross, you
can keep your beer. A nice vodka and tonic now, that’s more like it. Mmmmm.
Although I don’t think you could keep it cool enough in a camelback to be
really nice.

I think I’d have to be strapped to the lawnmower in some way to mow the lawn
though.