# Explain the logic behind proportions

Homework Helper
Gold Member
I'm trying to explain the logic behind proportions to someone who's just learning this stuff and find it very difficult to give a simple explanation.

exemple: A 46 cm tall gorilla weights 2kg. Approximately how much would king kong weight if he's 1800 cm tall?

How would you justify that the answer is 1800*2/46 ?

The justification is by saying that the gorilla weighs $$\frac{2}{46}\frac{kg}{cm}$$

I would say that this is the "average weight per unit". Once the person you are helping sees that proportion tell them to do the inverse, if the gorilla weighs $$\frac{2}{46}\frac{kg}{cm}$$, how much does he weigh if he is 46cm tall?

$$\frac{2kg}{46cm}*\frac{46cm}{1} = 2 kg$$

And then hopefully he/she can see the connection.

And then finally you can move on to the problem with 1800 cm.

Jameson

Last edited:
Prices usually help.

One oragne costs \$3. How much will five oranges cost? Then set it up as Jameson showed and it should click. Prices are things they deal with all the time.

Homework Helper
Gold Member
Two good ideas, thank.

honestrosewater
Gold Member
If they like geometry, you could use similar polygons. A few examples with rectangles and triangles should be enough.

HallsofIvy
Homework Helper
Quasar987 said:
exemple: A 46 cm tall gorilla weights 2kg. Approximately how much would king kong weight if he's 1800 cm tall?

How would you justify that the answer is 1800*2/46 ?
This is a really bad example! You can't justify that answer- it's wrong!

Weight is not proportional to height, it is proportional to body volume which is itself proportional to height cubed. The correct answer is (1800/45)3*2.

Homework Helper
Gold Member
If you want the postal adress of the authors of her textbook I can get it for you.

It IS a very bad exemple though. An 18m high King Kong that weights 78kg? Come on.

shmoe
Homework Helper
quasar987 said:
An 18m high King Kong that weights 78kg?
That's awesome. This is a good reminder that if you're trying to model some real world phenomena it's a good idea to do a sanity check on your proposed answer. If you end up predicting such a cream puff King Kong (how could this guy possibly go toe to toe with godzilla or with Fay Wray for that matter) then you might want to question the method used, especially if you've included it as a solution in a textbook.

A spherical balloon is partially blown up and its surface area is measured. More air is then added, increasing the volume of the balloon. If the surface area of the balloon expands by a factor of 9.4 during this procedure, by what factor does the radius of the balloon change?

If the radius of a sphere is increased by 12 %, by what factor does its surface area increase?
- By what percentage does its surface area increase?
- By what factor does the sphere's volume increase?
- By what percentage does the sphere's volume increase?

The weight of an object at the surface of a planet is proportional to the planet's mass and inversely proportional to the square of the radius of the planet. Jupiter's radius is 11 times Earth's and its mass is 320 times Earth's. An apple weighs 1.0 N on Earth. How much would it weigh on Jupiter?

I am sorry to resurrect this year old thread, but I am having trouble sanity-checking
I am developing an algorithm for assuring children that even if a 12 ft high spider did exist, it couldn't hurt you because it couldn't move.

Here is what I got stuck on:
The chart is based upon 72" high average person (index=me)
-------------------------------------------------------------------------------------------------------
h=Height k=new height w=weight y=h^3 w/y (expressed as a percent) = .0509
h in inches.......36........66........72........120...........1200
h^3............ 46656...287496..373248...1728000....17280000
w................23.75.......146......190........880...........879629
Using either my indexing model or
using the useful formula ((k/h)^3)*w from @HallsOfIvy above...
this model doesn't sanity-check. a 3' tall kid doesn't weigh 24 pounds

I guess I can get a very specific weight for a
100' tall exact copy of 'me'

Last edited:
HallsofIvy
Homework Helper

I am sorry to resurrect this year old thread, but I am having trouble sanity-checking
I am developing an algorithm for assuring children that even if a 12 ft high spider did exist, it couldn't hurt you because it couldn't move.

Here is what I got stuck on:
The chart is based upon 72" high average person (index=me)
-------------------------------------------------------------------------------------------------------
h=Height k=new height w=weight y=h^3 w/y (expressed as a percent) = .0509
h in inches.......36........66........72........120...........1200
h^3............ 46656...287496..373248...1728000....17280000
w................23.75.......146......190........880...........879629
Using either my indexing model or
using the useful formula ((k/h)^3)*w from @HallsOfIvy above...
this model doesn't sanity-check. a 3' tall kid doesn't weigh 24 pounds
Sigh! Once upon a time I was 66 inches high and weighed 146 pounds. Now I am 66 inches tall but much more than 146 pounds!

I guess I can get a very specific weight for a
100' tall exact copy of 'me'
According to this chart,
a 36" child might be anywhere from 25 to 75 pounds with the average a little below 50. Of course, children are not just "miniaturized" adults- they tend to be stockier and heavier for their height.

When talking about the ability to move, you should include in that that strength is proportional to length squared (because it depends on the cross section area of muscle fiber). A spider, blown up in length by a factor of 10 would be 100 times stronger but 1000 times heavier- so only 1/10 as able to support its own weight.

Last edited by a moderator:
CRGreathouse
Homework Helper

I'm trying to explain the logic behind proportions to someone who's just learning this stuff and find it very difficult to give a simple explanation.

exemple: A 46 cm tall gorilla weights 2kg. Approximately how much would king kong weight if he's 1800 cm tall?

How would you justify that the answer is 1800*2/46 ?
Ugh, what a terrible problem. Really, you'd expect weight to scale as the cube of height (or, at least, with an exponent between 2 and 3). 2*(1800/46)^3 would be more realistic... unless you think I weigh as much as King Kong.

When I left WV I was 72" high and 135 pounds. I have grown in cross-section

Thanks for the link, and the suggestion. I will post a link to the finished python code. *if it is ever finished...

Sigh! Once upon a time I was 66 inches high and weighed 146 pounds. Now I am 66 inches tall but much more than 146 pounds!

I guess I can get a very specific weight for a
100' tall exact copy of 'me'
According to this chart,
a 36" child might be anywhere from 25 to 75 pounds with the average a little below 50. Of course, children are not just "miniaturized" adults- they tend to be stockier and heavier for their height.

When talking about the ability to move, you should include in that that strength is proportional to length squared (because it depends on the cross section area of muscle fiber). A spider, blown up in length by a factor of 10 would be 100 times stronger but 1000 times heavier- so only 1/10 as able to support its own weight.

HallsofIvy
Homework Helper

Ugh, what a terrible problem. Really, you'd expect weight to scale as the cube of height (or, at least, with an exponent between 2 and 3). 2*(1800/46)^3 would be more realistic... unless you think I weigh as much as King Kong.
Yes, that was said in response #6, four and a half years ago! :rofl:

CRGreathouse