4th hour Polar project by Jennifer Swift

2019-04-24 20:51:08 UTC

2019-04-24 21:44:08 UTC

erios12 — 2019-04-26 14:54:28 UTC

2019-04-26 14:56:41 UTC

2019-04-26 14:57:56 UTC

erios12 — 2019-04-26 14:58:00 UTC

2019-04-26 14:58:01 UTC

syeom67 — 2019-04-26 15:00:16 UTC

2019-04-26 15:00:53 UTC

emayer11 — 2019-04-26 15:01:20 UTC

2019-04-26 15:01:33 UTC

2019-04-26 15:01:50 UTC

admitriev63 — 2019-04-26 15:02:19 UTC

agoode88 — 2019-04-26 15:02:44 UTC

2019-04-26 15:03:51 UTC

2019-04-26 15:05:01 UTC

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2019-04-26 15:45:34 UTC

2019-04-26 15:47:00 UTC

2019-04-26 15:47:19 UTC

progress

2019-04-26 15:51:15 UTC

admitriev63 — 2019-04-26 15:52:33 UTC

2019-04-26 15:52:54 UTC

2019-04-26 15:54:24 UTC

I changed my image

2019-04-26 15:56:09 UTC

work on original mage

emayer11 — 2019-04-26 15:57:27 UTC

2019-04-26 15:57:52 UTC

Progress 1

2019-04-26 16:01:31 UTC

agoode88 — 2019-04-26 16:01:37 UTC

2019-04-26 16:03:18 UTC

2019-04-26 16:06:16 UTC

2019-04-26 16:07:25 UTC

erios12 — 2019-04-26 16:10:01 UTC

2019-04-27 01:05:28 UTC

2019-04-27 01:35:53 UTC

2019-04-28 00:47:02 UTC

2019-04-28 21:27:44 UTC

2019-04-28 21:30:41 UTC

2019-04-28 21:30:52 UTC

work 4:30pm 04/28/19

2019-04-29 01:22:40 UTC

how to translate

jpetrofsky43 — 2019-04-29 16:18:14 UTC

1. write your polar function as f(theta)=yourfunction
you can restrict theta by adding {low<=theta<=high} after
2. replace f with a specific named variable for that function
you can make longer named variables by using _ after the first letter (such as l_eftflower)
3. make a new equasion r=f(theta)
(with f replaced with your variable)
this is the centered polar graph so you can see how it looks
4. YOU MUST USE RADIANS
paste these 4 equasions in:
\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}=f\left(\arctan\left(\frac{y-b}{x-a}\right)+\frac{\pi\left(1-\operatorname{sign}\left(x-a\right)\right)}{2}\right)
\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}=f\left(\arctan\left(\frac{y-b}{x-a}\right)+\frac{\pi\left(1-\operatorname{sign}\left(x-a\right)\right)}{2}+2\pi\right)
-\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}=f\left(\arctan\left(\frac{y-b}{x-a}\right)+\frac{\pi\left(1-\operatorname{sign}\left(x-a\right)\right)}{2}+\pi\right)
-\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}=f\left(\arctan\left(\frac{y-b}{x-a}\right)+\frac{\pi\left(1-\operatorname{sign}\left(x-a\right)\right)}{2}-\pi\right)
5. replace a and b with unique variables such as l_eftflowerX and l_eftflowerY and replace f with your other variable again
6. hide the other equations

jpetrofsky43 — 2019-04-29 17:49:55 UTC

2019-04-29 22:57:27 UTC

2019-04-30 16:39:27 UTC

2019-04-30 18:03:52 UTC

2019-05-01 15:06:51 UTC

2019-05-01 16:36:04 UTC

2019-05-01 16:39:30 UTC

some parts of it are screen shotted separately.

erios12 — 2019-05-01 17:14:30 UTC

2019-05-01 18:02:01 UTC

admitriev63 — 2019-05-01 21:14:24 UTC

2019-05-01 21:38:23 UTC

2019-05-01 22:21:54 UTC

2019-05-02 02:43:38 UTC

2019-05-02 12:52:20 UTC

2019-05-02 13:02:16 UTC

work 04/30/19

2019-05-02 13:30:52 UTC

syeom67 — 2019-05-02 16:35:00 UTC

2019-05-02 19:47:48 UTC

2019-05-03 00:52:00 UTC

2019-05-03 15:06:41 UTC

2019-05-03 15:07:43 UTC

2019-05-03 15:09:39 UTC

2019-05-03 15:10:04 UTC

2019-05-03 15:17:26 UTC

2019-05-03 15:28:52 UTC

2019-05-03 15:34:23 UTC

Final

2019-05-03 19:54:13 UTC

2019-05-03 20:26:09 UTC

2019-05-03 20:26:25 UTC

agoode88 — 2019-05-05 02:41:52 UTC

2019-05-05 16:46:28 UTC

Final Product

2019-05-05 17:27:21 UTC

2019-05-05 17:39:37 UTC

2019-05-05 18:17:35 UTC

2019-05-05 19:52:38 UTC

2019-05-05 23:59:35 UTC

2019-05-06 01:57:56 UTC

emayer11 — 2019-05-06 02:07:41 UTC

emayer11 — 2019-05-06 02:08:10 UTC

emayer11 — 2019-05-06 02:08:30 UTC

jenniferswift4_1 — 2019-05-10 13:21:17 UTC

Graded - No progress

jenniferswift4_1 — 2019-05-10 13:30:19 UTC

Graded

jenniferswift4_1 — 2019-05-10 13:30:44 UTC