# Thread: Golden Ratio Based from 1..10 ?

1. ## Golden Ratio Based from 1..10 ?

My wife and I were having this discussion on the golden ratio being used to measure beauty and thought it would be fun to calculate our own ratings based 1..10 (10 being the perfect golden ratio 1.618)

We have each of our scores see below:

first face mesaurment (height / width) = 1.571
second face mesurment (height / width) = 1.666

How can we calculate a number based on 10, with 10 being the perfect golden ratio 1.618?

We tried the followig but it doesn't seem correct:
first face:
golden ratio (1.618 - 1.571) = 0.047
10 - 4.7 = 5.3 ???
second face:
golden ratio (1.618 - 1.666) = -0.048
10 - 4.8 = 5.2 ???

Thanks,
-Peter

2. Let $\varphi$ be the golden ratio, and let $f$ be a face ratio. Then $0\le|f-\varphi|<\infty$. A natural idea is to have a linear function that maps the difference $|f-\varphi|$ to [0, 10] such that the difference 0 is mapped to 10. However, one has to select what value is mapped to 0, i.e., which face ratio is absolutely ugly. For example, one can decide that the ratio 3 is absolutely ugly, so $3-\varphi$ is mapped into 0. The function that maps $[0, 3-\varphi]$ into [10, 0] is $g(x)=10-10x/(3-\varphi)$, so the measure of beauty for a face ratio $f$ is $g(|f-\varphi|)=10-|f-\varphi|/(3-\varphi)$. Good news: for your values, the measures are both around 9.97; this is because they are very far from the ugliness standard.

3. Originally Posted by emakarov
Let $\varphi$ be the golden ratio, and let $f$ be a face ratio. Then $0\le|f-\varphi|<\infty$. A natural idea is to have a linear function that maps the difference $|f-\varphi|$ to [0, 10] such that the difference 0 is mapped to 10. However, one has to select what value is mapped to 0, i.e., which face ratio is absolutely ugly. For example, one can decide that the ratio 3 is absolutely ugly, so $3-\varphi$ is mapped into 0. The function that maps $[0, 3-\varphi]$ into [10, 0] is $g(x)=10-10x/(3-\varphi)$, so the measure of beauty for a face ratio $f$ is $g(|f-\varphi|)=10-|f-\varphi|/(3-\varphi)$. Good news: for your values, the measures are both around 9.97; this is because they are very far from the ugliness standard.
I'm a bit rusty with my math but it seems as though the"ugliness" ratio is subjective so would it make more sense to re-write the equation and base it on the perfect golden ration 1.618? If so what would the equation look like? My wife and I are still in disagreement over this.

Thanks

4. Originally Posted by psmith05
I'm a bit rusty with my math but it seems as though the"ugliness" ratio is subjective so would it make more sense to re-write the equation and base it on the perfect golden ration 1.618? If so what would the equation look like? My wife and I are still in disagreement over this.

Thanks
I believe the least subjective mapping to the scale [0;10] involves psychiatrists and public surveys...
Otherwise you can only say difference (or difference in percentage) and be precise about that.

On the other hand, emakarov provided you with a very thorough example of a simple mapping. If that method satisfies you, my personal advise would be to measure an ugly face ratio and replace number 3 in emakarov's function with your desirable ugly-face-measurement.

5. it seems as though the"ugliness" ratio is subjective so would it make more sense to re-write the equation and base it on the perfect golden ration 1.618?
If you want to come up with a grade from 0 to 10, then you need to know what faces earn a 0 (the other end of spectrum is agreed on: golden ratio is worth a 10).

6. Originally Posted by emakarov
Let $\varphi$ be the golden ratio, and let $f$ be a face ratio. Then $0\le|f-\varphi|<\infty$. A natural idea is to have a linear function that maps the difference $|f-\varphi|$ to [0, 10] such that the difference 0 is mapped to 10. However, one has to select what value is mapped to 0, i.e., which face ratio is absolutely ugly. For example, one can decide that the ratio 3 is absolutely ugly, so $3-\varphi$ is mapped into 0. The function that maps $[0, 3-\varphi]$ into [10, 0] is $g(x)=10-10x/(3-\varphi)$, so the measure of beauty for a face ratio $f$ is $g(|f-\varphi|)=10-|f-\varphi|/(3-\varphi)$. Good news: for your values, the measures are both around 9.97; this is because they are very far from the ugliness standard.
Thanks emakarov,

Like I said both my wife and I are rusty with our math. When I use your formular to solve for g I keep getting 182.36 so obviosuly I'm missing something. I've gone over the equation several times. Here is how I arrived at my answer.

Using the face ratio of 1.571 I took the absolute value of the difference from the left hand side g(.047)
After calculaing the right side of the equation I got 9.953/1.382 = 8.571
To solve for g I divided 8.571/0.47 and got my 182.36

Any idea where I went wrong?

Thanks,
Peter

7. Originally Posted by psmith05
Thanks emakarov,

Like I said both my wife and I are rusty with our math. When I use your formular to solve for g I keep getting 182.36 so obviosuly I'm missing something. I've gone over the equation several times. Here is how I arrived at my answer.

Using the face ratio of 1.571 I took the absolute value of the difference from the left hand side g(.047)
After calculaing the right side of the equation I got 9.953/1.382 = 8.571
To solve for g I divided 8.571/0.47 and got my 182.36

Any idea where I went wrong?

Thanks,
Peter
g(|1.571 - phi|) = g(0.047) = 10 - |1.571 - phi|/(3 - phi) = 10 - 0.047/1.382 = 10 - 0.034008683 = 9.965991317

Like I said before, you might want to replace (3 - phi) with more realistic (desirable-ugly-face-measurement - phi) in the formula as face ratio of 3 corresponds to face like this (see below) lol

8. Using the face ratio of 1.571 I took the absolute value of the difference from the left hand side g(.047)
After calculaing the right side of the equation I got 9.953/1.382 = 8.571
To solve for g I divided 8.571/0.47 and got my 182.36
Here g is not a number. This is how I called a function that maps an interval [1.382, 0] into [0, 10]. So, $g(1.382) = g(3 - \varphi) = 0$ and $g(0) = 10$.

Speaking about this, I realized I made an error. I first wrote that $g(x)=10-10x/(3-\varphi)$, but then said that the final grade is $10-|f-\varphi|/(3-\varphi)$. Of course, it should be $10-10|f-\varphi|/(3-\varphi)$, or $10(1-|f-\varphi|/(3-\varphi))$. This bumps down your grade from 9.97 to about 9.66.