What size ( in inches) is a 108" girth box?
PLEASE do not tell me how to measure a box to get to 108"!
Please give me an example of a maximum box size.
For example, a 30’ X 20" X 20" would equal 108". Any rectangularish shaped box sizes.
I’m trying to send as large a box as possible, I DO NOT have a box right now to measure!
I will buy one if I can just find out what maximum size I can send through the mail is.
Looks like boxes can be bought in 28" x 28" x 12" sizes. That would be the 108" maximum that USPS accepts, if I understand their measurement procedures correctly.
It’s not 108" girth, it’s girth + length can’t exceed 108". Assuming you meant 30", and not 30’, your example would equal 110" (20*4 = 80 + 30 = 110). They define length as the longest dimension and girth as the distance around the thickest part. So for a cubical box the maximum size would be 21.6" on all sides. Iggy’s example works, but so would 28x20x20.
In case it’s not obvious from what Doug said, it’s the longest measurement plus twice the sum of the two shorter measurements.
If the dimensions are L" x W" x H" where L is the largest number, the limit is L+2W+2H = 108.
And don’t over pack the box, especially wrt padding, such as wadded paper and bubble wrap. The counter person will measure the girth with a flexible tape, and bulgy sides can add an inch or two to the girth.
Dan
We could ask what the largest possible volume you can send through the US Postal Service is. This means we want to maximize the quantity W * H * L subject to the restriction L + 2W + 2H = 108.
The result, if I’ve done my math correctly, is that an 18" x 18" x 36" box has the largest possible volume that complies with regulations (6.75 cubic feet). The box size Iggy found has a volume of 5.44 cubic feet, and the cubical box proposed by DougK has a volume of 5.83 cubic feet.
Of course, you might not be able to fit what you want to send into a box of those dimensions, but it’s still a fun calculus exercise.
A cylinder has a girth of its circumference, by USPS rules. That turns out to optimize to the same length as your calculation, 36" (and a circumference of 72"), but the volume is greater at 8.594 ft[sup]3[/sup].
Step by step instructions here.
Closest approximation to a sphere is 4 equal sides.
108/4 = 27.
You probably cannot buy 27" boxes.
Find the most cubical box you can, that matches your desired girth.
I was curious why the answer for the length seemed to be the same for either a circular or square cross-section, and saw that a triangular cross-section has the same answer as well (36" length). So I did the math:
We can express the volume as thus:
l * C(L - l)(L - l)
Where:
l: length of box
L: total length+girth by USPS rules (108")
C: A constant that converts a circumference squared into an area. This is always possible: a square would have C=1/16 (since a 1x1 square has a circumference of 4, and 44/16=1; a circle has C=1/(4pi)).
Multiplying out, we get:
C(lL[sup]2[/sup] - 2Ll[sup]2[/sup] + l[sup]3[/sup])
To find the maxima, we set the derivative to zero:
C(L[sup]2[/sup] - 4Ll + 3l[sup]2[/sup]) = 0
We note first that the value of C is irrelevant; it divides out completely. C came from the shape of the cross-section but doesn’t appear in the answer. Beyond that, we simply plug into the quadratic formula, and get two answers: L and L/3. L is a local minima (and a volume of 0). L/3 is the answer we’re looking for; 108"/3=36", which is the same as what MikeS and I came up with earlier.
Under the USPS rules a cube would need to be 108" / 5 = 21.6" sides.
The issue is that the USPS “length times girth” formula does not measure volume, presumably it’s a proxy that they use because it involves only addition but no multiplication, and maybe it correlates well to how easy it is to move something around.
So the question is how to maximize volume while satisfying the USPS formula. As explained by MikeS & Dr Strangelove, the solution is not a cube.
The largest volume for surface is a cube. Thus, L+2W+2H=108, W=L, H=L
5L=108 implies L=21.6"
Your best bet is a cube 21.6 inches on a side for maximal volume.
No, the solution for maximizing volume under the USPS restriction is not a cube because the USPS formula does not measure volume.
No. MikeS had it right. If the restriction were L+W+H=108", then you would be right, but it’s L+2W+2H=108", therefore W and H are more heavily weighted. So the optimal solution would have a smaller W and H than L. Though H and W would still be equal, because those are weighted the same.
Volume of such a box is V = L * W * H = L * W^2 = (108 - 2W - 2H) * W^2 = (108 - 4W) * W^2.
Take a derivative: dV/dW = -12 (W-18) W
Obviously this becomes zero at w=18, which means the volume of the box is maximized at W = 18". So the largest volume rectangular box for L+2W+2H=108 is 36"x18"x18".
(And who says calculus isn’t useful??)
What I said was a bit off - if the USPS gave a volume limit directly, then of course there would be no problem to solve in maximizing volume.
scr4 stated the situation more accurately, that the USPS formula is based on a sum of linear dimensions, but does not weight the linear dimensions equally, and that’s why the solution is not a cube.
Doh!