A Fermi estimate of vegetarianism on an energy basis


The other day, I was at a lakehouse without any means to correct to the Internet and was discussing vegetarianism as something desirable to save the planet (considering that meat is expensive to produce on the environment). Here's how I argued that it is, based on limited information.

First step is to pick something to compare the choices to one another. Ideally, we'd want something that can be compared on some level, so comparing self-hunted wild game to hydroponic bell peppers might not quite cut it. Besides, it needs to be something on which enough data is known: I know very little about how much wild game might reasonably ingest per time unit, for example. One thing we have much information on, between my relatives and I, is garden-grown crops and livestock, so I settled on comparing ground beef to romaine lettuce.

First, we need to estimate how large a grazing space would be needed to feed a single cow in a year. My relatives estimated around 1 acre (which is 300 feet x 300 feet, which we approximate to 100 meters x 100 meters). This makes some sense to someone who has seen what volume of hay is contained in a square meter of field. This assumes that hay is collected once a year which is not completely unreasonable in a place with seasons such as Canada.

Then, we estimate that a cow equals about 700 lbs of usable meat (based on my relatives experiences of getting cows butchered). Based on my experience with mass cooking, 3 lbs of ground beef is about the amount of meat that goes into a pâté chinois (traditional meal made from ground beef, potatoes and corn) which will last for 10 meals. This is a very conservative estimate since we don't include the potatoes and corn in the calculations.

700 lbs / 3 lbs is ≈250. 250 * 10 meals is 2500 meals. Excluding breakfast (as it is uncommon to eat meat for breakfast here), we need 365 days/year * 2 meals/day which comes out to about 700 meals/year. 2500 meals / 700 meals/year is about 3.5 years (21/7 = 3, 28/7 = 4).

On to romaine lettuces. Assuming that one plant can be grown in about 30 cm x 30 cm, we'll say that's 3 plants/m (well, technically, it should be 3 sqrt(plant)/m or 9 plants/m^2). So our acre gives us 3 plants/m × 100 meters = 300 plants. 300 × 300 is 90000 plants. Based on my relatives's estimates, a lettuce head can be harvested once every month for 5 months in Canada (the head is cut from the roots when harvested and can be grown back), but we'll ignore this like we did for the hay.

We now suppose that the equivalent of 2 full meals, on an energy basis, is about 10 lettuce plants. So we get 9000 days worth. Even approximating 1000 days/year, we still get 9 years worth of food. Using a closer approximation of 333 days/year gives us 27 years or 9 years at 30 lettuce plants per day.

Suffice to say that the argument in favor of vegetarianism is well supported by a back of the enveloppe calculation.

But does the estimate actually hold up?