math nerds

kuzi16

i have a problem for you.

i am looking for a generic equation to solve this problem:
you have two different sized boxes. you wish to stack them next to each other and have the tops of the box align and be level.
there is one catch...
one of the stacks is on a platform.

x = height of the fist box
y = height of the other box
a = height of the platform
H = height of the finished stacked boxes
for this problem lets assume that the boxes of height "y" are on the platform and that you have an infinite number of both boxes. the platform is immobile and neither of the boxes can change in size.


if there was no platform then the problem would be simple: xy=h (7 x 4 = 28 (for instance))

EDIT: im looking for the equation that will give you H, not the equation that will give you the number of each boxes needed.

urbino

H = x(y+(a/x))

For example, if the platform were 6" high:

H = 7(4+(6/7)) = 34

Or that's my guess, anyway.

Dustin1981

Those are the crazy equations I was getting as well. I think it will change each time depending on the platform height.

phobicsquirrel

urbino:

H = x(y+(a/x))

For example, if the platform were 6" high:

H = 7(4+(6/7)) = 34

Or that's my guess, anyway.

It's been far too long for me.. lol. Might help with all the nuts I have to organize...

michaelharvey

just a thought if the tops are going to be aligned wouldn't the hight of one of the boxes be the same as the finished stack. for example X (being the taller box) would be equal to H.

Garen B

I'll check it out a bit more when I wake up tomorrow but I believe this will work for you:

H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

Fourtotheflush

Garen B:

I'll check it out a bit more when I wake up tomorrow but I believe this will work for you:

H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

Thats a good one. Is this real or hypothetical?

kuzi16

Garen B:

I'll check it out a bit more when I wake up tomorrow but I believe this will work for you:

H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

since some multiple of x = h at some point, wont the equation be cx = dy + a ?
where "c" and "d" are the number of boxes needed from each? or am i now throwing in new variables that will throw everything off?

if it were h = x(y+a) then you would be adding the height of the platform every for every box "y"

kuzi16

Fourtotheflush:

Is this real or hypothetical?

i was stacking boxes at work earlier and one was on a platform. i didnt need them to be level but my mind was wondering. this is where it wondered to.

Fourtotheflush

kuzi16:

Fourtotheflush:

Is this real or hypothetical?

i was stacking boxes at work earlier and one was on a platform. i didnt need them to be level but my mind was wondering. this is where it wondered to.

Urby's formula looks good, im not sure it is spot on, but I will start cracking on it. I didnt get much sleep last night so I may not get back to you in the AM.

Garen B

I figured it out and checked it a few times to make sure.

Let k = constant of box 1, J = constant of box 2, a = height of platform and x be the variable.

k(j+a) + S{(kx)(jx)

The first part of the equation is to get both box k and box j + platform a to a level surface, then the second part is the summation of k and j to any multiple of x. What this means is that this equation is good for finding any height where both boxes are level.
Here's an example using numbers:

k = 5, j = 4, a = 3
5(4+3) + S (5 x 2)(4 x 2)
35+80 = 115 inches.
Checking the answer: 115/5 = 23, 115 - 3(To account for platform height) = 112/4 = 28

Fourtotheflush

Garen B:

I figured it out and checked it a few times to make sure.

Let k = constant of box 1, J = constant of box 2, a = height of platform and x be the variable.

k(j+a) + S{(kx)(jx)

The first part of the equation is to get both box k and box j + platform a to a level surface, then the second part is the summation of k and j to any multiple of x. What this means is that this equation is good for finding any height where both boxes are level.
Here's an example using numbers:

k = 5, j = 4, a = 3
5(4+3) + S (5 x 2)(4 x 2)
35+80 = 115 inches.
Checking the answer: 115/5 = 23, 115 - 3(To account for platform height) = 112/4 = 28

Phew, saved me the ice cream headache! We used to do problems like this in college all the time.

urbino

kuzi16:

Garen B:

I'll check it out a bit more when I wake up tomorrow but I believe this will work for you:

H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

since some multiple of x = h at some point, wont the equation be cx = dy + a ?
where "c" and "d" are the number of boxes needed from each? or am i now throwing in new variables that will throw everything off?

if it were h = x(y+a) then you would be adding the height of the platform every for every box "y"

For every box "x", you mean. Yeah, that's how it looks to me, too. I think the same would apply to geebee's second, way cool looking formula.

Your formula works, I think, but you'd have to know both c & d, in addition to x, y, and a, to find out the H you were looking for. If my formula works -- and I think it does, but I was never all that great at math -- you only need to know x, y, and a, the heights of the 3 objects.

cabinetmaker

My formula is to take a skillsaw, remove the shelf, then stack the boxes in an alternating pattern. Thinking outside the - wait for it -










BOX.

jsnake

ok, I have tolerated political talk and religious talk but I refuse to sit here and let this mathematics talk keep going. This is wrong.

kuzi16

urbino:

kuzi16:

Garen B:

I'll check it out a bit more when I wake up tomorrow but I believe this will work for you:

H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

since some multiple of x = h at some point, wont the equation be cx = dy + a ?
where "c" and "d" are the number of boxes needed from each? or am i now throwing in new variables that will throw everything off?

if it were h = x(y+a) then you would be adding the height of the platform every for every box "y"

For every box "x", you mean. Yeah, that's how it looks to me, too. I think the same would apply to geebee's second, way cool looking formula.

Your formula works, I think, but you'd have to know both c & d, in addition to x, y, and a, to find out the H you were looking for. If my formula works -- and I think it does, but I was never all that great at math -- you only need to know x, y, and a, the heights of the 3 objects.

in a real life situation you would know the x,y,and a. you plug those in to my formula and you plot them on a graph. this will give you the c and d for every combo. ... or thats what i was thinking till i read this:

Garen B:

I figured it out and checked it a few times to make sure.

Let k = constant of box 1, J = constant of box 2, a = height of platform and x be the variable.

k(j+a) + S{(kx)(jx)

The first part of the equation is to get both box k and box j + platform a to a level surface, then the second part is the summation of k and j to any multiple of x. What this means is that this equation is good for finding any height where both boxes are level.
Here's an example using numbers:

k = 5, j = 4, a = 3
5(4+3) + S (5 x 2)(4 x 2)
35+80 = 115 inches.
Checking the answer: 115/5 = 23, 115 - 3(To account for platform height) = 112/4 = 28

serioulsy dude.... you are a nerd.

on that note, can you explain to me why yours works and mine wouldnt? (i have to admit, i didnt test my formula)

Garen B

I'm studying to be an engineer so I'll take that as a compliment.

The equation works because it's basically doing what your equation does but has that first part in there to compensate for the platform. That way you'll start off with a fresh, level surface before finding other heights that satisfy your requirements of being level.

Think of it as making a building, you pour a good solid foundation so that you can get as level as possible and then build up from there.

phobicsquirrel

all that time figuring out a formula, you could have done it.

Lasabar

The only math I worry about is when I have to calculate ounces in multiples of 12

j0z3r

Lasabar:

The only math I worry about is when I have to calculate ounces in multiples of 12

Huh...here I thought you just counted the cans.