iamcal.com Since May 2000

• Happy new year!
• Newbie topic - May 2016
• Newbie topic - April 2016
• Newbie topic - March 2016
• Newbie topic - February 2016
• Newbie topic - January 2016
• Newbie topic - November 2015
• Newbie topic - August 2015
• Newbie topic - July 2015
• Newbie topic - May 2015
• Newbie topic - April 2015
• Newbie topic - March 2015
• Newbie topic - September 2014
• Newbie topic - December 2014
• Newbie topic - October 2014
• Newbie topic - August 2014
• Newbie topic - July 2014
• Newbie topic - June 2014
• Newbie topic - May 2014
• Hello!

As you seemed to enjoy the last maths question so much, here is one for me. This is a bit more 'triangle' orientated (which is what I am good at, so it's what you get =] ).

Assume you have 2 3D map references, points A, B and C, and wanted to travel to points X, Y, and Z.

What is the equation for the shortest distance between the references?

say what?

what

Don't worry boys, just leave it to cal

easy. get the intermediate triangular distance:

i = square_root(((a-x)^2) + ((b-y)^2))

then use it with the third dimension:

d = square_root((i^2) + ((c-z)^2));

so the whole thing:

d = square_root((square_root(((a-x)^2) + ((b-y)^2))^2) + ((c-z)^2));

I was never good at math.

O righto?!

you're just showing off now..

ok, can you draw a diagram?

yes

Close, but not quite...

We can prove it by plugging some numbers into your equation.

For ease, we'll make a=0, b=0, and c=0, and x=-10, y=-20, z=-30.

so:

d = square_root((square_root(((10)^2) + ((20)^2))^2) + ((30)^2))

calculate

d = square_root(square_root((100 + 400)^2) + (900))

sore more

d = square_root(square_root(250000) + (900))

more

d = square_root(square_root(250900))

more

d = square_root(500.899)

finally

d = 22.38

Now, by my reckoning, that makes the distance between the two points less than the distance between c and z, which can't be true...

My head hurts reading that :(

ZZZZZZZZZZ

sqrt((sqrt(10^2+20^2))^2+30^2) =37.4Is that the same formula? not sure.EDIT: I think there was a bracket in the wrong place?.EDIT#2: Oh. There isn't. Hmm.

'Scuse the 'drawn with a mouse' look.

izb, did you draw that with a mouse?

i did.

MS Paint?

Paint shop pro. The mathematician's choice.

MS Paint?

Is that a repackaged but fundamentally similar version of paint, with the only difference being a pink bow on the head of the brush??

wokka-wokka-wokka-wokka

yeah, you use pythagoras theorem.
(i wanted to put that to prove that im not pants at maths, but i cant be arsed to solve it as above.)

if grim could actually do the working without making any mistakes then you'd see that i am correct:

d = square_root((square_root(((a-x)^2) + ((b-y)^2))^2) + ((c-z)^2));

d = square_root((square_root(((10)^2) + ((20)^2))^2) + ((30)^2));

d = square_root((square_root(((100) + (400))^2) + (900));

d = square_root((square_root(500^2) + 900);

d = square_root(500 + 900);

d = square_root(1400);

d = 37.416573867739413855837487323165

what did the constipated mathematician do?

work it out with a pencil!

an oldy and a goody

Ah! I think I misunderstood your brackets for in step 5 it is the root of 500^2 rather than the root of 500^2 + 900.

Gotcha.

Yeah, that's the same as what I had, but in a different order.

Part two of the maths challenge comes tomorrow!

did anyone take part in the maths challenge at school? it was a multiple choice exam and they used to tell you not to guess the answers because you got marks deducted for wrong answers. the first two years i did it properly but never finished so i never got an award. the third year i guessed every single answer on the paper.. i got a bronze award!! a guy in my class got a gold award and came sixth in the country or something ridiculous...

Yeah I did it. Can't remember if I got a gold but I definatly got a silver one year.

I expect cal invented teh Math's challenge when he was 4 and set all the questions each year until he was 11 blindfold and with he arms tied behind his back.

yeah, i did it.
my claim to fame for that was getting the school's highest standardised score when i was in year 8 or something.
i always managed better in the science / chemistry / computing competitions.

i've never heard of such a thing

but i was pretty good at maths at school

Okay, here comes part two of the challenge!

Using the equation below (or above, we did it differently to how it is done in my Maths Challenge book), and assuming an accelleration rate of 1 meter per second.

If the object had to move from a standstill at point A and come to a standstill at point B, where would it be at any given point in time?

Now I don't have the answer to this, and I haven't figured it out yet.

I know the equation for distance travelled at any time, but I can't figure out what grid reference the point will be at, and that's what we need to know.

If its a wednesday, it would be queueing up to sign on.

is this related to the first question?

is point A a,b,c and B x,y,z?

were the distances before in metres?

Yeah, point A is a,b,c and point B is x,y,z.

The distances can be in any unit, as you are looking for an equation.
Metres would be simplest, reckon.

the distances can't be in just any unit if the acceleration is in metres per second!

Grim, you're not developing weapons of mass destruction are you?

Remember the launch of the PS2 when the word went round that Iraq had bought 20,000 to control Nasty weapons?

Er, okay, I guess it must be metres.
As long as the accelleration unit and the mesurement unit match, it could be anything from milimetres to light years, I suppose...

i got 3 bronzes and a silver in maths challenge. woo.

I've got it but it's messy.

Basically, using our old equation to get the distance - let's assume that we are travling from X12, Y44, Z366 to X0, Y0, Z0.

The distance equation coughs up 368.83 (r).

So (ugly version) X12 Y44 Z366 = 368.83

Divide by X (12)

X1 Y3.6 Z30.5 = 30.73 (We'll call this distace D)

It's easy to work out how far we have travled if we know the accelleration. Assuming 1m/s, after five seconds we moved 15 meters.

Divide 15 by D (30.73) = 0.48 (r).

Now we have a distance;

0.48(x, y, z)

or

0.48(1, 3.6, 30.5)

or

1.48, 1.72, 14.64

So, after 5 seconds we have moved that far in each direction.

Adjust our original grid reference by the above

(12, 44, 366) - (1.48, 1.72, 14.64)

And we find our new references:

(X10.52, Y42.28, Z351.36)



Phew. Someone who can be arsed can do the equation, but I think that's right.

It's easy to work out how far we have travled if we know the accelleration. Assuming 1m/s, after five seconds we moved 15 meters.

hmm. starting from zero, after 5 seconds we'll be going at 5m/s. so or average speed for 5 seconds was 2.5m/s so we've actually gone 12.5 metres.

i haven't really read the rest of your solution.

fuck knows