Surface area and length (percent increase relationship)

In summary, the conversation discusses the relationship between the percentage increase in length and the percentage increase in surface area for an object. It is stated that the surface area increase should be in line or less than the length increase. However, there is some debate about this statement and whether it is always true, with different scenarios and examples being discussed. Ultimately, it is concluded that for small changes in size, the percentage increase in surface area may not always be in line with the percentage increase in length.
  • #1
simito_
2
0
TL;DR Summary
Surface area and length increase
Hi all,

While calculating the surface area for an object, I was told the below statement. However, I am not sure is this correct, please can someone help me to explain this with an example? Is the below statement always true?

The surface area % increase should be in line or less than the % length increase.

Thanks,
 
Physics news on Phys.org
  • #2
simito_ said:
While calculating the surface area for an object, I was told the below statement. However, I am not sure is this correct, please can someone help me to explain this with an example? Is the below statement always true?

The surface area % increase should be in line or less than the % length increase.
Yes, it is a correct statement. [Edit: Providing that we are scaling only "length" and not "width" and "height" as well].

For a flat object such as a sheet of paper, you can see that the surface area will be directly proportional to the object's length, right?

For an object with some extent in the third dimension, things are trickier. You can see that the cross-sectional areas of the top and bottom will have increased. The cross-sectional areas of the front and back (looking perpendicular to the length axis) will have increased as well. But the cross-sectional areas of the two ends will have remained unchanged. This is admittedly a bit hand-wavy because the surface area of a shape is not equal to the sum of the cross-sectional areas of its six sides.There is a more rigorous way to prove the inequality. I am not sure that I can do justice to the reasoning in a short post...

Suppose that we imagine the surface of the unstretched object tiled with a bunch of rectangular panes. Maybe you fill in the gaps with strange shapes. But it is clear that you can tile essentially the whole shape with rectangular panes if you make them small enough -- you just use smaller and smaller panes as you fill in whatever irregular gaps remain. [We are doing a surface integral].

We make sure that all of the panes are lined up with a "width" axis that is perpendicular to the direction of stretch. The "length" axis for each pane will be more or less parallel to the direction of stretch.

If it helps, imagine a spherical Earth being stretched by pulling the north and south poles apart. We divide the surface into north-south strips. The strips would be wedge shaped, so we trim to make them rectangular and cover the places where we trimmed with smaller rectangles, repeating the process indefinitely.

When we stretch the object, the width (east/west) of each of the panes will remain unchanged.

When we stretch the object, the length (more or less north/south) of each of these panes will increase by at most the percentage of stretch. Near the poles, there will be negligible increase. Near the equator there will be the full increase.

So the surface area of each pane will increase by at most the percentage of stretch. So the total surface area of the object will increase by at most the percentage of stretch.
 
Last edited:
  • Like
Likes russ_watters and simito_
  • #3
Welcome to PF.

simito_ said:
The surface area % increase should be in line or less than the % length increase.
I do not believe that is the case for changes in more than one dimension.

For small changes of size;

If you have a two-dimensional object and scale all linear dimensions by 1%, the surface area will scale by 2%, because area is the square of the scale.

If you have a three-dimensional object and scale all dimensions by 1%, the surface area will scale by 2%, while the volume will scale by 3%, because volume is the cube of the scale.
 
  • Like
Likes jbriggs444, russ_watters and simito_
  • #4
@jbriggs444 seems to be reading "length increase" as a stretch in one dimension, whereas @Baluncore I think is reading it as a linear scale factor. So starting with a 1×1×1 cube and applying a "length increase" of 2, @jbriggs444 gets a 2×1×1 cuboid while @Baluncore gets a 2×2×2 cube.

Perhaps @simito_ could clarify which scenario is intended.
 
  • Like
Likes russ_watters, jbriggs444 and simito_
  • #5
@jbriggs444 @Baluncore and @Ibix

Thanks for the responses.

I will give you an example here:

Lets assume I have a product made of different complex shapes and there is wire at the end of the product which is 180cm, Now what i am trying to do is to add 20 cm length (made of different shape). Based on my calculation,

Percent Increase in length is 11.11%
Percent Increase in surface area is 11.54%

Note - 19.22 cm^2 is the LSA for product A, all i did is added a length of 20 cm with surface area 2.2192 cm^2.Based on my calculation here, Percent Increase in length is not >= Percentage Increase in surface area. However, is the vise versa.

DescriptionPercent Increase in length isPercentage Increase in surface area
PRODUCT A=200 MM/ 1800 *100
=11.11%
2143.91-1922=221.93 mm^2
221.93/1922*100 =11.54%
Product A+ extra length (20 cm)
2143.91

Thanks,
Smitha
 
  • #6
simito_ said:
Lets assume I have a product made of different complex shapes and there is wire at the end of the product which is 180cm, Now what i am trying to do is to add 20 cm length (made of different shape). Based on my calculation,
This sounds like neither myself nor @Baluncore understood your scenario.

You are not maintaining the same shape and scaling up one or more dimensions. You are instead adding an additional blob on one side.

There is no hard and fast rule for how the surface area of the boundary will change as a result. It could increase greatly, increase slightly, stay the same or even decrease.

Example: A sphere with a divot. Fill in the divot and the surface area decreases.

Example: A sphere. Tack on a fluffy folded flag at the north pole and surface area increases greatly.
 
Last edited:
  • Like
Likes simito_ and phinds
  • #7
jbriggs444 said:
There is no hard and fast rule for how the surface area of the boundary will change as a result. It could increase greatly, increase slightly, stay the same or even decrease.
I agree.

Due to the lack of a clear 1D, 2D or 3D specification in the OP, my post #3 took the opportunity to counterpoint the 1D interpretation in post #2, with the 3D interpretation.

And now I get to mention "Small is beautiful" by E. F. Schumacher.
https://en.wikipedia.org/wiki/Small_Is_Beautiful

The 1% general scaling rule, in our real world:
2% is the area, cost of painting, or the insulation of a 3D object.
3% is critical to volume, mass and the cost of a 3D object.
 
  • Like
Likes simito_, jbriggs444 and russ_watters

FAQ: Surface area and length (percent increase relationship)

What is the relationship between surface area and length in geometric shapes?

The relationship between surface area and length in geometric shapes depends on the dimensions of the shape. Generally, for a given shape, if the length increases, the surface area increases, but not necessarily at the same rate. For example, in a cube, if the length of a side doubles, the surface area increases by a factor of four (since surface area is proportional to the square of the length).

How do you calculate the percent increase in surface area when length changes?

To calculate the percent increase in surface area when the length changes, you first need to know the initial and final lengths. Calculate the initial and final surface areas using the appropriate geometric formulas. Then, use the formula: Percent Increase = [(Final Surface Area - Initial Surface Area) / Initial Surface Area] * 100%.

Why does the surface area increase faster than the length?

The surface area increases faster than the length because surface area is a two-dimensional measure, while length is a one-dimensional measure. For many shapes, surface area is proportional to the square of the length. Therefore, any change in length results in a squared change in surface area, leading to a faster increase.

Can you provide an example of how surface area and length percent increase relationship works in a cube?

Sure! Consider a cube with an initial side length of 2 units. The initial surface area is 6 * (2^2) = 24 square units. If the side length increases to 4 units, the new surface area is 6 * (4^2) = 96 square units. The percent increase in surface area is [(96 - 24) / 24] * 100% = 300%.

Does the percent increase in surface area always follow a predictable pattern with length for all shapes?

While the percent increase in surface area with respect to length follows a predictable pattern for many regular geometric shapes (e.g., cubes, spheres), the specific relationship can vary for more complex or irregular shapes. For regular shapes, the surface area typically scales with the square of the length, but for irregular shapes, the relationship may not be as straightforward and may require more complex calculations or empirical data.

Similar threads

Replies
4
Views
744
Replies
33
Views
2K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
4
Views
1K
Back
Top