the Otherpower.com Discussion Board || Comments Hydro in the Gutters

Front Page - [Homebrewed Electricity-- (wind) (solar) (hydro) (steam) (controls) (storage) (mechanical)] - Classifieds - Site News
Everything - Newbies - [Remote Living-- (housing) (heat) (light) (water)] - Rants & Opinion - Diaries - Our Products

Hydro in the Gutters | 67 comments (67 topical, 0 editorial)

Re: Hydro in the Gutters (3.00 / 0) (#23)
by MattM on Fri Jun 13th, 2008 at 08:35:42 PM MST
(User Info)

Using math we can figure the time necessary for the water to fall 75 feet by using the equation for the distance that the object falls in a given time. This equation is d = gt²/2

where
    d is the distance the object falls
    g is the acceleration due to gravity
    t is the time considered
    t² is t times t or t-squared
    gt²/2 is g times t-squared divided by 2

so 75 feet = (32.17 feet/seconds²)t²/2

    (32.17 feet/seconds²)t² = 150 feet
    t² = 150 feet / 32.17 feet/seconds²
    t² = 4.6627293 seconds²
    t = 2.1593354 seconds

feet / 2.1593354 seconds = 34.732909 feet / second
333 pounds / second * 34.732909 feet / second = 602 foot-pounds per second, which is also 1.09 horsepower.

Correct my math where it went wrong.
----------------------------- Go Huskers!
[ Parent ]

Hydro in the Gutters | 67 comments (67 topical, 0 editorial)

Menu

· create account
· How to use the board
· FAQs
· search the board
· Google search the board
· Old Otherpower Board

Login

Make a new account

Username:

Password:

You must be a registered user to post here. It's easy and free, and the link is on the upper right side of your page.
All trademarks and copyrights on this page are owned by their respective companies. Postings are owned by the poster, but may be deleted or moved at the ADMIN's sole discretion. The Rest © 2003 Forcefield.
You can Email the board ADMIN here. PLEASE include the username you signed up with!