Thanks for contributing both to this community and to the academic study of art.

Beautiful selection to highlight on your post.

What if this was the equivalent of like a 17th century shit post?

Sexual repression’s a helluva drug

They were definitely geeking out on something! I had this piece of visionary poetry in my dissertation:

I entered - where - I did not know,

Yet when I found that I was there,

Though where I was I did not know,

Profound and subtle things I learned;

Nor can I say what I discerned,

For I remained uncomprehending,

All knowledge transcending.

• John of the Cross, Verses Written on an Ecstasy.

NB "ecstasy" as we know it did not exist in the 16th Century 😂

Hey my man gotta explain the white stains on his shirt to the church !

The most underrated movie ever.

how does this baby look like its 50?

i think it's miraculous laminar flow

Check out some ancient pottery, bath walls, and a lot of art to see that most sex acts and kinks have been round for millenia.

Rule 34 was really just copypasta to start with.

From ChatGPT:

Me: Water exits a nozzle at coordinate (0, 1.7m) and hits the ground at (2m, 0). Assume the water exits the nozzle at 90 degree (i.e. straight to the right). How can we determine the speed of the water?

(here I made some rough assumptions based on the image, and assumed that her feet were at (0, 0)

Answer: The speed of the water exiting the nozzle can be determined using the principles of projectile motion.

1. Determine the time of flight:

   The water is projected horizontally from a height of 1.7 meters and hits the ground at 2 meters horizontally. The vertical motion is governed by the equation: [ y = y_0 + v_{y0} t - \frac{1}{2} g t^2 ] Here, ( y_0 = 1.7 ) meters, ( v_{y0} = 0 ) m/s (since there is no initial vertical velocity), and ( g = 9.8 ) m/s² (acceleration due to gravity).

   The water hits the ground when ( y = 0 ): [ 0 = 1.7 - \frac{1}{2} \times 9.8 \times t^2 ] Solving for ( t ): [ 1.7 = 4.9 t^2 ] [ t^2 = \frac{1.7}{4.9} ] [ t = \sqrt{\frac{1.7}{4.9}} \approx 0.59 \text{ seconds} ]

2. Determine the horizontal velocity:

   The horizontal distance covered is 2 meters. Since horizontal velocity ( v_x ) is constant (there is no horizontal acceleration), the distance can be given by: [ x = v_x t ] Here, ( x = 2 ) meters and ( t \approx 0.59 ) seconds: [ v_x = \frac{2}{0.59} \approx 3.39 \text{ m/s} ]

Therefore, the speed of the water exiting the nozzle is approximately 3.39 m/s.