Shot Photography Contest Pictures - Image Page 10

Moon Shot - created by bf2015

Rank: 1/5

Score: 67.5% (72)

11568 views

So this is actually an HDR pano of the moon shot with a Canon 550D through a Meade LX90 8" telescope. It has a fixed F-stop of F10. The equivalent DSLR lense would be right around 2300mm. It was taken just before sunset.

6 shots total:
-1EV/0EV/+1EV
HDR through Photomatix Pro 3
Stitched in Photoshop CS5 (5 years and 3038 days ago)

1 Comment

Moon, Shot

iC0DY says:

beautifully done.

Howdie stranger!
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The Money Shot - created by locksmagic

Favs: 0

SBS: n/a

Hi-res

Rank: 5/5

Score: 64.8% (0)

6455 views

A pitcher stands less than 20 meters from the batter, and throws the ball at speeds of around 40 meters/sec. In that time the ball drops from its "straight line" path by about 1.2 meters.

Because of the stitches on the baseball there is considerable turbulence affecting its motion, and it may behave somewhat erratically. The complications in the motion of the ball arise from the Prandtl Layer of air that moves with the baseball, and which has a dominating effect on the motion of the ball.

The pitcher may put considerable spin on the baseball, corresponding to about 20 revolutions over its path.

A batter has about 0.5 seconds to gauge the path of a pitch, and the bat must be at the right place [within about 1 cm] at the right time [within 0.01 seconds] to make solid contact.

The bat-ball collision lasts typically for 0.001 seconds, and the average force on this ball is of order 10,000 Newtons, corresponding to a mass of about 1000 kg [or a ton].

Taking this into consideration if a batter manages to make contact with s baseball that is pitched between 30-45 m/s has the potential to leave the bat at a speed of 60 m/s spinning up to 3000 rpm and can return to the mound in 0.45 sec.

If the baseball traveling at a potential speed of 45 m/s contacts a stationary object i.e. a camera some of the kinetic energy is transferred to the stationary object. Kinetic energy is equal to Â½ m v 2. The transfer of energy in this case will be equal to 1/2 the ball's mass multiplied by the square of it's velocity.

The amount of potential damage that a camera would incur if impacted by a by a baseball is incalculable as there are too many variables to account for. Therefore attempting to capture an image such as this could be considered by some to be gutsy...

Hopefully this explanation will appease the mod who told me I needed to add a description. :-D (5 years and 3047 days ago)