First Principle Calculations for Land Required to Feed One Person

Insolation is 1000 W/M2. To achieve 100W of food growth - efficiency of capturing sunlight is 10% - and conversion of this captured energy to plant matter is another 10%. So 1% of solar energy is converted to plant matter. 1 square meter gives us 10W of food production. Thus, minimally, 10 square meters get us to the 100W level. Assuming that 50% of the best photosynthesizer can be eaten, such as algae, that is 200 square feet as the theoretical minimum land required to feed one person.

A person uses about 100W - [1] or 8M Joules per day.

Potatoes produce 40,000 lb per acre in Idaho - [2]. There are 350 Calories in a pound of potatoes - [3]. A person needs 6 lb of potatoes per day. That means 2000 lb per year, or 1/20 of an acre. 1/20 of an acre is 2,200 sf. If aquaponics produces 10x the yields of standard production, then we can consider 220 sf as the area requirement, based on first principle concepts.

So how to guarantee this in a greenhouse? You have to eat your lawn! How? Indirectly via Oyster mushrooms or bunnies. But, there are only 159 Calories in a pound of Oyster mushrooms - [4].

A rabbit pair can produce 600 lb of meat per year - and 4 lb of feed are required to produce one lb of meat - [5]. 1 kg of rabbit meat requires 4 minutes of production time - [6]. And giant rabbits can weigh up to 25 lb - [7]. They take 10 weeks to reach 5 lb [8] and have a dressing out percentage of over 50% [9], and a pound of rabbit is 800 calories.

All right, we need to go further. 1 egg is 78 calories - [10].

Add one pound of Tilapia - 435 Calories. [11].