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. That is 100 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 calculations.

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

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].

All right, we need to go further. 1 egg is 78 calories - [9].

Add one pound of Tilapia - 435 Calories. [10].