Photo-A-Day layout math - DPChallenge Forums

Canon EF 17-40mm f/4L USM

01/06/2005 02:00:20 PM · #1

I need your help:

(2x+a)*(2y+b)-(a*b)=365

Also, let (2x+a)/(2y+b)=3/2

y=?
x=?

or

a=?
b=?

If possible, let a/b=(2x+a)/(2y+b) which should mean that a/b=3/2 ...otherwise ignore this part.

Message edited by author 2005-01-06 14:01:10.

kirbic

Canon EOS R5

Canon RF 24-70 f2.8 L IS

01/06/2005 02:13:20 PM · #2

You've got two equations and four unknowns. In order to solve, you'd need four equations. If we let a/b = 3/2, then we still need one more equation.

Jacko

Canon EOS-20D

01/06/2005 02:15:30 PM · #3

You might want to refer your problem to Laurieblack . She loves math.

Canon EF 24-70mm f/2.8L USM

01/06/2005 02:20:10 PM · #4

Originally posted by kirbic:

You've got two equations and four unknowns. In order to solve, you'd need four equations. If we let a/b = 3/2, then we still need one more equation.

Okay...did I do this right?

(2x+a)*(2y+b)-(a*b)=365
(2x+a)*(2y+b)=365+(a*b)
2x+a=[365+(a*b)]/(2y+b)

x={[365+(a*b)]/(2y+b)-a}/2

and likewise for y?

This might help me...

ButterflySis

Canon EOS-50D

Canon EF 50mm f/1.8 II

01/06/2005 02:22:08 PM · #5

It's been so long since I've done this stuff. I get this far (so far)...

(2x+a)*(2y+b)-(a*b)=365

multiply it out to get...

4xy+2xb+2ay+ba-(a*b)=365

2y(2x+a)+(b(2x+a)-(a*b)=365

The two paranthesis are the same...and I forget what you do with them. Do you remember?

Anyway, you said...

a/b = 2x+a)/(2y+b) = 3/2

So substitute that into what you have. Like...

2y(3)+b(2x+a)-(a*b)

And keep going...?

I dunno, maybe I've confused you more. :-)

Jen

ButterflySis

Canon EOS-50D

Canon EF 50mm f/1.8 II

01/06/2005 02:22:59 PM · #6

Originally posted by thatcloudthere:

Originally posted by kirbic:
You've got two equations and four unknowns. In order to solve, you'd need four equations. If we let a/b = 3/2, then we still need one more equation.

Okay...did I do this right?

(2x+a)*(2y+b)-(a*b)=365
(2x+a)*(2y+b)=365+(a*b)
2x+a=[365+(a*b)]/(2y+b)

x={[365+(a*b)]/(2y+b)-a}/2

and likewise for y?

This might help me...

I think you're right. I made it too complicated. Good work!

doctornick

Apple iPhone 13 Pro

01/06/2005 02:23:46 PM · #7

Originally posted by Jacko:

You might want to refer your problem to Laurieblack . She loves math.

Yeah, she's the math wiz of the site! She'll solve that in no time!

Message edited by author 2005-01-06 14:24:08.

Canon EF 70-200mm f/2.8L USM

01/06/2005 02:25:50 PM · #8

Actually, it's unsolvable...kirbic's right.

It's like asking if I have 50 pennies and buy an apple, how many pennies do I have left? oh, and how much was the apple?

hsteg

Fujifilm X-T20

01/06/2005 02:26:54 PM · #9

Originally posted by doctornick:

Originally posted by Jacko:
You might want to refer your problem to Laurieblack . She loves math.

Yeah, she's the math wiz of the site! She'll solve that in no time!

for real. shes good at math.

Canon EF-S 55-250mm f/4-5.6 IS

01/06/2005 02:34:31 PM · #10

Here's the practical version of my question:

I want to organize my 365 photo-a-day pictures in a rectangle (each photo will take a up a square space). The red part will have 365 photos. The yellow part will have text.

How many photos wide should it be, and how many photos high? I want an even number of photos above and below, as well as to the right and left of the yellow space. I want both to have an approximate 3:2 w:h aspect ratio...

Canon EOS-60D

01/06/2005 02:37:22 PM · #11

Originally posted by ButterflySis:

OK... picking up where Jen left off...

2y(3)+b(2x+a)-(a*b)...

2 years (really 3) of math + b (like that grade would ever happen) (2x dropping the class + a (I was really dreaming then) - (a (no chance in hell) * b (that won't happen either) = Pick A Different Major! :o)

01/06/2005 02:49:53 PM · #12

Okay, it's just not possible...I'm convinced! I've been trying (using Excel) to work this out and it won't work.

paulmath

Canon EOS-20D

01/06/2005 02:51:06 PM · #13

No sweat:

a=3
b=2
x=(-12+sqrt(144+4*6*365))/8
y=(-12+sqrt(144+4*6*365))/12

Other solutions are possible.

Canon EF 70-200mm f/2.8L IS USM

01/06/2005 02:55:32 PM · #14

I guess my logic was flawed when I presented the equation...I need all whole numbers.

My practical presentation of the problem is more accurate...

...see the pretty picture below...

Message edited by author 2005-01-06 14:56:09.

scottwilson

Canon EOS-350D Rebel XT

Canon EF 28mm f/2.8

01/06/2005 03:01:06 PM · #15

y =-b/2-sqrt(730+2*a*b)/16 or -b/2+sqrt(730+2*a*b)/16
x = -a/2+2*sqrt(730+2*a*b) or -a/2-2*sqrt(730+2*a*b)

EddyG

Canon EOS-1D Mark II

01/06/2005 03:10:07 PM · #16

Since 365 is an odd number, I don't think there is a way to divide it into an even number of rows and columns.

There are several ideas that might work well, however. One idea is to make it 12 rows tall (one for each month), and have 31 columns (for each day of the month). For the months with 30 (or 28) days, you can insert a "panoramic" shot (or three, for February) to make it "fit". If you stagger these wide shots throughout the design, it could look quite good.

The other option would be to just leave "non-picture spaces" at the end of each row. You could put small amounts of text in those squares, or just leave them "empty". This would provide a more "calendar" feel, since you'd be able to look at a particular square and now exactly what day it was taken on...

01/06/2005 03:14:46 PM · #17

Originally posted by thatcloudthere:

I guess my logic was flawed when I presented the equation...I need all whole numbers.

You need whole even numbers as well, otherwise it will be lopsided. Also, you can't get exactly 365, as it's not even.

It's easier to work it out as a large rectangle AB with a smaller rectangle CD removed from the middle, with the unit being one photo.

If the aspect ratio of the piece is the same as the photo (like, 3:2), then you can simply further with A=B and C=D (or approx equal).

The closest I got was 20x20 - 6x6 = 400-36 = 364, so you miss a day. But there is probably a better solution.

01/06/2005 03:32:47 PM · #18

If you wait for a leap year, or leave a square blank.

24x24 - 14x15 = 366
33x34 - 27x28 = 366
36x36 - 30x31 = 366
39x39 - 33x35 = 366
42x44 - 38x39 = 366

Canon EF 28-135mm F/3.5-5.6 IS USM

01/06/2005 03:40:32 PM · #19

Originally posted by mickwest:

Originally posted by thatcloudthere:
I guess my logic was flawed when I presented the equation...I need all whole numbers.

You need whole even numbers as well, otherwise it will be lopsided. Also, you can't get exactly 365, as it's not even.

It's easier to work it out as a large rectangle AB with a smaller rectangle CD removed from the middle, with the unit being one photo.

If the aspect ratio of the piece is the same as the photo (like, 3:2), then you can simply further with A=B and C=D (or approx equal).

The closest I got was 20x20 - 6x6 = 400-36 = 364, so you miss a day. But there is probably a better solution.

This is exactly what I was trying...and that's as close as I got! Nothing else will work if I'm bent on doing it that way. I'll have to think of other ways...I like EddyG's idea, but I would almost rather not divide it by month if possible.

You could also try some stuff with 377 squares (365 + 1 square for the beginning of each month that just says "January" or whatever)...I don't think that helps, though...

dsa157

Canon EOS-50D

01/06/2005 03:41:38 PM · #20

Originally posted by mickwest:

The closest I got was 20x20 - 6x6 = 400-36 = 364, so you miss a day. But there is probably a better solution.

thatcloudthere never explicitly stated it, but if he is going to put this on a web page, he should rethink the design anyway. Even high bandwith visitors will not want to wait for 365 (more really) individual HTTP requests to return to load a page of thumbnails.

Dave

Canon EOS-60D

Canon EF-S 55-250mm f/4-5.6 IS

01/06/2005 03:41:39 PM · #21

I'll just hide in the corner until all this computation is over... ;o)

smokeditor

Canon EOS-5D Mark II

Canon EF 24-70mm f/2.8L USM

01/06/2005 03:43:30 PM · #22

im doing mine on a 20x 30 print and I am makeing 16x23 squares that are 1.125 inches each. That makes 368 squares, so 3 of the photos will take up 2 squares. This also allows me to have a 1 inch border on the top and sides and 3.125 inchs on the bottem to title text

01/06/2005 03:43:55 PM · #23

For even numbers you need to go to 368 photos:

20x20 - 8x4 = 368
24x26 - 16x16 = 368
26x28 - 18x20 = 368

01/06/2005 03:49:20 PM · #24

Originally posted by dsa157:

You could just use an image map and a single image, if this is indeed a problem.

See: //www.htmlgoodies.com/tutors/cs_imap.html