Optimize Your Rooftop Space with Aquaponics - A Step-by-Step Guide

  • Thread starter Thread starter utsharpie
  • Start date Start date
AI Thread Summary
The discussion focuses on optimizing rooftop space for aquaponics using gutters filled with water to grow hydroponic basil. The setup involves two 10'X20' gutters, each weighing 650 lbs, placed on a sloped roof with a U-shaped design. The user seeks advice on how to support this weight without drilling into the roof, potentially using the wedge formed by the sloping rooftops. The design aims to distribute the weight evenly over a 400 square foot area while connecting the gutters to a counterbalance system. The goal is to create a sustainable rooftop garden without compromising the structural integrity of the home.
utsharpie
Messages
9
Reaction score
0
first, this is for my actual home's rooftop and isn't school related. I am not in a physics class.

i have 2 areas of 10'X20' of gutters filled halfway with water on my inclined roof. the gutters are 10 feet long and have a radius of 2inches. each gutter is evenly placed one foot apart from the next as you go down from the top of the roof towards the ground. there are 2 of these 10'X20' each weighing 650lbs with combined plant weight side by side on a roof top that is 45'X25'.

the gutters will be connected to a counterbalance on the otherside of the rooftop. the distance from the point of the roof that the counterbalance can hang is a max of 2 feet. the roof top is 25 feet from the edge closest to the ground to the top of the roof and the height of the rooftop to the bottom edge of the roof closest to the ground is 4feet. i guess the angle is 35 to 40 degrees
 

Attachments

  • rooftop2.GIF
    rooftop2.GIF
    6.4 KB · Views: 524
Physics news on Phys.org
roof angle as wedge

could i some how use the wedge where 2 sloping rooftops come together as a wedge?

if you are looking down onto the roof from straight overhead the roof is a U shape and each of the three slopes towards the center, so as if you were looking into an upside down hollow pyramid(make sense?). and i am trying to support 650 lbs of evenly distributed weight over a 400 square foot area. could i use the shape of the roof to somehow wedge support the weight without having to drill into the roof?

please any help is appreciated for design ideas.
 
utsharpie said:
first, this is for my actual home's rooftop and isn't school related. I am not in a physics class.

i have 2 areas of 10'X20' of gutters filled halfway with water on my inclined roof. the gutters are 10 feet long and have a radius of 2inches. each gutter is evenly placed one foot apart from the next as you go down from the top of the roof towards the ground. there are 2 of these 10'X20' each weighing 650lbs with combined plant weight side by side on a roof top that is 45'X25'.

the gutters will be connected to a counterbalance on the otherside of the rooftop. the distance from the point of the roof that the counterbalance can hang is a max of 2 feet. the roof top is 25 feet from the edge closest to the ground to the top of the roof and the height of the rooftop to the bottom edge of the roof closest to the ground is 4feet. i guess the angle is 35 to 40 degrees

Is that on top of the roof? On the outside? What is that to be used for?
 
yes on the roof top. i will be growing hydroponic basil on my roof NFT style using gutters as the plant troughs.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top