Square footage conundrum

[bkreigsr]

If I buy a shed with a floor that is 10 x 5, that's 50 sq ft, right?

Now if i take 2 feet out of the base and transfer it to another shed, and add 2 feet to the side and do the same transfer - the new shed becomes 12 x 3, correct, that's 36 sq ft, right?

Somehow with the equivalent dimensions, I've lost 14 sq ft ?

This has to do with our local construction permit regulations and the maximum size you can build without a permit.

The dimensions in the example are hypothetical for the sake of simplicity.

Do I belong back in 5 th grade?

Bill K

[unclebilly]

The best thing I can suggest is to get some graph paper (the stuff with lines spaced at 1/4” in up and across) and sketch out both scenarios.

Use each grid as a square foot. Count the squares for each scenario.

You will quickly see what is going on and want to delete this thread.

[stomachmonkey]

Whatever x’s whatever is this many Whatever’s

10
10
10
10
10

12
12
12

Each sq ft already covers length and width, the math is just how many you have in a row and how many rows

[Dantilla]

Correct.

In both, you have a total of 30 feet of wall.
5+10 + 5+10=30
12+3 + 12+3=30

The most area possible would be a square. As the shape gets longer and skinnier, area inside diminishes.
Eventually you would have two 15 foot walls with zero space between/inside.

[masraum]

Originally Posted by unclebilly

The best thing I can suggest is to get some graph paper (the stuff with lines spaced at 1/4” in up and across) and sketch out both scenarios.

Use each grid as a square foot. Count the squares for each scenario.

You will quickly see what is going on and want to delete this thread.

Originally Posted by Dantilla

Correct.

In both, you have a total of 30 feet of wall.
5+10 + 5+10=30
12+3 + 12+3=30

The most area possible would be a square. As the shape gets longer and skinnier, area inside diminishes.
Eventually you would have two 15 foot walls with zero space between/inside.

If you're moving around bits of wall, that's completely different from moving around tiles from the floor. The scenario that Dantilla describes above is more like where you can afford X number of feet of fence and you want to fence in an Area with that same length of fence. In that case, like he says, the more square you can get the building the greater the area will be inside. As he's said, a square building will have more sq footage than a building with the same amount of wall that's long and narrow.


If one shed is 10x5, and you make it 10x3, you're removed 20sqft. If you take that 20 and add it to another shed to make it 10x7, then you've increased that shed from 50 to 70 sqft. If you increased the other side so the new shed is 12x5, then you only added 10sqft, but you've got 10 floor tiles left.

[jyl]

What are the local regs?

Originally Posted by bkreigsr

If I buy a shed with a floor that is 10 x 5, that's 50 sq ft, right?

Now if i take 2 feet out of the base and transfer it to another shed, and add 2 feet to the side and do the same transfer - the new shed becomes 12 x 3, correct, that's 36 sq ft, right?

Somehow with the equivalent dimensions, I've lost 14 sq ft ?

This has to do with our local construction permit regulations and the maximum size you can build without a permit.

The dimensions in the example are hypothetical for the sake of simplicity.

Do I belong back in 5 th grade?

Bill K

[BK911]

Originally Posted by bkreigsr

If I buy a shed with a floor that is 10 x 5, that's 50 sq ft, right?

Now if i take 2 feet out of the base and transfer it to another shed, and add 2 feet to the side and do the same transfer - the new shed becomes 12 x 3, correct, that's 36 sq ft, right?

Somehow with the equivalent dimensions, I've lost 14 sq ft ?

This has to do with our local construction permit regulations and the maximum size you can build without a permit.

The dimensions in the example are hypothetical for the sake of simplicity.

Do I belong back in 5 th grade?

Bill K

I was told there would be no maths.
This isnt a 5th grade problem, really a calculus problem.
Max/mins are determined by a derivative.

Want to maximize area wrt length of sides:

Optimization eq; A = LxW
Constraint eq; P = 2L + 2W or W = (P-2L)/2

So A = (PL)/2 -L^2

Take derivative of A WRT L and set to zero

0 = P/2 - 2L or P = 4L

So whenever all 4 sides are equal, you have maximized the area.

Or to take back to 5th grade level:

L and W = 10; P = 40, A = 100
L=8 W=12; P=40, A = 96


This is why I built my retirement cabin in a square.
Maximize living space with minimized roofing, flooring, etc...

[masraum]

Originally Posted by BK911

I was told there would be no maths.
This isnt a 5th grade problem, really a calculus problem.
Max/mins are determined by a derivative.

Want to maximize area wrt length of sides:

Optimization eq; A = LxW
Constraint eq; P = 2L + 2W or W = (P-2L)/2

So A = (PL)/2 -L^2

Take derivative of A WRT L and set to zero

0 = P/2 - 2L or P = 4L

So whenever all 4 sides are equal, you have maximized the area.

Or to take back to 5th grade level:

L and W = 10; P = 40, A = 100
L=8 W=12; P=40, A = 96


This is why I built my retirement cabin in a square.
Maximize living space with minimized roofing, flooring, etc...

Correct, calculus. Also correct, equal length sides will give greatest area. (assuming what you're concerned with is the amount of wall that you have to work with).

The most efficient way to enclose a space (with 4 square sides) is via a square (or cube for 3D). That way you get the max amount of square footage or volume for the minimum amount of wall/roof/floor.

I haven't ever checked, but I suspect that a circle/sphere would be even more efficient.

[wilnj]

Originally Posted by masraum

I haven't ever checked, but I suspect that a circle/sphere would be even more efficient.

Thats correct. A square with an area of 100sf would have a perimeter of 40lf.

A circle with an area of 100sf would have a circumference of 35lf.


Sent from my iPhone using Tapatalk

[BK911]

Shoot, I should have built in a circle?
Too late now.

[Dantilla]

Originally Posted by BK911

Shoot, I should have built in a circle?

A circle with a circumference of 30 feet would have 72 square feet of area.
71.78 if one cares about sig figs.


circle: 72 Sq Ft
square: 56 Sq Ft.
Rectangle: Less and less Sq Ft as we go further away from square.

What should masraum do with all the extra floor tiles?

[masraum]

Originally Posted by Dantilla

A circle with a circumference of 30 feet would have 72 square feet of area.
71.78 if one cares about sig figs.


circle: 72 Sq Ft
square: 56 Sq Ft.
Rectangle: Less and less Sq Ft as we go further away from square.

What should masraum do with all the extra floor tiles?

Use them instead of baseboard or maybe make a nice backsplash?

[71T Targa]

As noted, the square gives the largest footprint with the least wall space.

And I know this because I helped my 5th grader with her math last year and this the exact problem we worked on. (With a fence, but same)

[bkreigsr]

Originally Posted by Dantilla

Correct.
In both, you have a total of 30 feet of wall.
5+10 + 5+10=30
12+3 + 12+3=30
The most area possible would be a square. As the shape gets longer and skinnier, area inside diminishes.
Eventually you would have two 15 foot walls with zero space between/inside.

So 30 ft of wall in each shed does not translate to equal areas?
I realize these responses have been dumbed-it-down a bit (a lot) for me, but my head still hurts like someone is squeezing it in a vise, trying to make my eyes pop out of the sockets.
Bill K

[masraum]

Originally Posted by bkreigsr

So 30 ft of wall in each shed does not translate to equal areas?
I realize these responses have been dumbed-down a bit (a lot) for me, but my head still hurts like someone is squeezing it in a vise, trying to make my eyes pop out of the sockets.
Bill K

Here's a picture. You can count the squares.


All of these are 30 units worth of wall in different configurations.

You can see that the long skinny version (14x1) is only 14 square units of area. The next is 12x3 like the question, and therefore only 36 square units of area. Then we have the 10x5 for 50 square units of area and finally 7x8 (almost square) for 56 square units of area.

A circle that's got a 20 foot diameter is going to be about 314 square feet inside. A circle with a 28 foot diameter is going to be about 628 square feet inside. So a 40% larger radius will yield an area twice as large (at that size).

A cube that's 1 foot on all sides will be 1 cubic foot. A cube that's 2 feet on all sides is 4 cubic feet, and a cube that's 4 feet on all sides is 64 cubic feet. So double the sides (1' -> 2') and you quadruple the volume. Double the sides again (2'->4') and you increase the volume by 16 times.

[bkreigsr]

Thanks Steve.

[rusnak]

Build a geodesic dome, and keep the 70s alive.

[RWebb]

for the 2 sheds, I assume you will build them so that a foundation or footer can be added in between them, and a roof built over the fnd, and maybe walls too

you know, when the code changes to make it legal

[Tom '74 911]

But... what color is the shed going to be?

(I'm an architect and this is the only thing that most of my clients and HOA boards would care about...)

: ) : )

[Bill Douglas]

Originally Posted by bkreigsr

and the maximum size you can build without a permit.

Find out if you can build the shed to the maximum size AND have a veranda/porch roof. then when it's signed off and the neighbors have stopped spying on you, you move the front wall out to enclose the veranda.