# Overview

Consider the following system for sustainable food production: The purpose of this system is to provide an environmentally-friendly, semi- (or fully-) automated means to produce food for lots of people. Similar in nature to the Earthship question, but delving into a few more details.

# Problem

I need help performing calculations on multi-storey indoor farms, using vertical plan production. I'm trying to reconcile the following variables:

• People to feed.
• Building dimensions and storeys.
• Total wattage necessary for lighting.

For example:

• 1,000 people
• 64m x 64m x 6 storeys x 4m per storey
• 7.71MW

# Calculations

The following sections serve as an example calculation.

## Energy

For the purposes of an example, assume a population of 350 people (but it could be 1,000 or any other number). The minimum global daily intake as of 2006 is 1850 calories per person, which is 7740kJ per person.

• daily = 7740kJ
• input = 350 x 7740kJ x 365 = 988785MJ

For a given population of 350 people, they require a minimum of 988785MJ of energy per year.

## Plant Growth

Since we know the energy per year needed by the population, it is possible to determine the size and harvests of a food that will yield that energy amount.

A giant large red round heirloom tomato stores 85kJ of energy per 100g, and reaches an average of 375g in 65 days, which equals 318kJ. (Round the number of annual harvests down to 5 to err on caution.)

• energy = 318kJ (per tomato)
• yield = 50 (per plant)
• harvests = 5 (per year)

The number of plants is calculated as:

• plants = input / (harvests x yield x energy)
• plants = 988785MJ / (5 x 50 x 318kJ)
• plants ~= 12450

The space required for each tomato plant:

• diameter = 2m
• height = 1.5m

## Building Area

The minimum building area is calculated using:

• area = (pi x diameter/2) x plants
• area = (pi x 2m^2/2) x 12,450
• area = 3.14159265m2 x 12,450 = 39,113m2

Since indoor farms use vertical space, we can pick an arbitrary number of storeys for the building. In this example, we'll choose 10. At 10 storeys, each storey must occupy 3,911.3m2. Therefore, a square building would be approximately 63m per side.

Using plants 1.5m tall, means a storey 3m tall fits two levels of plants per storey. This halves the building length and width to 31.5m per side. We still need to account for interstitial space for infrastructure (e.g., ducts, plumbing, wiring). This adds 10.5m per side, bringing the total length and width to ~42m per side. Thus:

• storeys = 10
• plant floor space = 31.5m x 31.5m
• interstitial = 1/3 (10.5m)
• total floor space = 42m x 42m

## Lighting

Using high-efficiency LED grow lamps:

• racks = 2
• coverage = 1.2192m2
• lamps1 = plant floor space / coverage x racks x storeys
• lamps1 = 992.25m2 / 1.2192m2 x 2 x 10
• lamps1 ~= 16277

Using light movers increases coverage by 35 per cent, therefore:

• lamps2 = lamps1 - (0.35 x lamps1) ~= 10,581
• movers = lamps1 - lamps2 = 5,696

## Energy Requirements

The electrical needs are calculated as:

• power = 250W (per lamp)
• total1 = lamps2 x power ~= 2.65MW
• total2 = total1 + movers x 5W = 2.68MW

My primary concern is whether total1 is correct.

Plasma lighting seems unbelievably efficient, which could reduce substantially reduce power requirements even further.

Nate Storey calculated the following:

• Using current LED efficiencies.
• Sustain (calorie + nutrition) 1,000 people on a variety of foods.
• The building is 64m x 64m (approx. 1 acre) by 6 storeys.
• Each storey must be 4m (approx. 13 feet) high.
• Vertical plane production within the vertical building.
• Requires 45 watts/sq. ft. per hour, at 18 hours per day for ~3.5 square feet of production.

I think this implies:

• area = 64m x 64m x 6 = 24576m2 area
• power = 45W/ft2 = 484.376W/m2
• energy = area x power = 24576m2 x 484.376W/m2 = 11.9MW

His calculations show ~11.9MW to feed 1,000 people; my calculations show ~2.68MW to feed 350 people, which is ~7.66MW to feed 1,000.

# Questions

What assumptions made for total1 are incorrect, if any, and how can they be corrected?

• You haven't reference any of your sources. I think the core problem is that both sets of numbers are very soft. When you start with "I assume this, give or take 50%" and proceed to layer more assumptions like that on top, getting out the end only 40% different is sheer luck. Actual farmers who've been doing this for decades can't get the accuracy you seem to expect.
– Ⴖuі
Nov 16, 2015 at 8:18
• Of the first three numbers one is an assumption (350 people), the other two need references (7500kJ x 365). You apparently require 3 significant figures (better than .5% accuracy), rounding off the length of a year to 365 gives you a starting error of -0.07%, about 1/10th of your total allowed error. So a reference saying "both caclulations use a 365 day year" would be useful. I think you'd best specify timezone/localtime as well, because half a day is another 2/10ths of your allowed error. You're asking someone to do a lot of work, but you haven't demonstrated that you'd appreciate the result
– Ⴖuі
Nov 18, 2015 at 6:13
• You should also be aware that you have long chains of calculations, with several parameters in between that have uncertainties (error bars). We know from numerical analysis that e.g. calculating X [+/- dX] times Y [+/- dY] gives us as XY [+/- dZ], where the error dZ derives from dX and dY, but I forgot the exact calculation (last time I did that is approx 30 years ago). So the more assumptions in your chain, the larger the uncertainty in the final outcome.
– user2451
Nov 18, 2015 at 11:31