Sustainable Vegitarianism

September 19th, 2006-09-18                                          Phil Mackenzie
 
Assignment 2 : ENVE3000 – Dr. Hart
       
        Assumptions:
 

1. The meat comes from the wild or from an industry which does not make use of the designated agricultural land.
2. The meat eater population is of the third trophic level (1st level Carnivores) and consumes primary consumers from the second trophic level (Herbivores).
3. One tenth of one percent of the energy coming from the sun is absorbed by the primary producer trophic level (Autotrophs).
4. A maximum of one tenth of the energy absorbed by the Autotrophs can be absorbed by the Herbivores.
5. A maximum of one tenth of the energy consumed by the Herbivores can be absorbed by 1st level Carnivores.

        Givens:
 
1. Daily Human Food Energy Requirement: 1000 Cal/day
2. Solar Energy Hitting the Land: 403 cal/cm2/day
3. Area of the Entire Island: 5000 km2
4. 50% of the land is used for agriculture
 
        Question:
 

1. What population level can be sustained with a food source of 100% vegetables.
2. What population level can be sustained with a food source of 75% vegetables and 25% meat.

Answer:

 
First, assign the units so that the unit of Energy is uniformly expressed in kiloCalories (Cal)
 
403 cal/cm2/day = 403*10-1 Cal/cm2/day
 
Then, express all area in km2
 
(403E-1 Cal/cm2/day) * 1000002 cm2/km2 = 403E7 Cal/km2/day
 
Find the area being used for agriculture, this area will produce a crop of autotrophs which absorb the sun-energy.
 
50% of 5000 km2 = 0.5 * 5000 km2= 2500 km2
 
1)    Let Vmax represent the maximum number of vegetarians to live on the island.
 
Vmax * 1000 Cal/day = 403E7 Cal/km2/day * 2500 km2* 0.1 (loss from 1 trophic level) * 0.001 (loss from solar absorption percentage)
Vmax  = (100.75E7 Cal/day) / (1000 Cal/day)
 
Vmax  = 1 007 500 people
 
 
2)    Let Mmax represent the maximum number of people who eat 25% meat and 75% vegetables
 
Mmax * 1000 Cal/day= 0.25*[403E7 Cal/km2/day * 2500 km2* 0.01 (loss from 2 trophic levels) * 0.001]
                                    + 0.75*[403E7 Cal/km2/day * 2500 km2* 0.1 (loss from 1 trophic level) * 0.001 ]
Mmax = (780812500 Cal/day) / (1000 Cal/day)
 
Mmax = 780 812 peoplespan lang=”EN-CA” style=”font-size: 10pt; font-family: Verdana;”> * 0.1 (loss from 1 trophic level)                                        * 0.001 (loss from solar absorption percentage)

Vmax  = (100.75E7 Cal/day) / (1000 Cal/day)

Vmax  = 1 007 500 people

2)    Let Mmax  represent the maximum number of people who eat 25% meat and 75% vegetables

Mmax * 1000 Cal/day= 0.25*[403E7 Cal/km2/day * 2500 km2 * 0.01 (loss from 2 trophic levels)                                        * 0.001 (loss from solar absorption percentage) ]

                              + 0.75*[403E7 Cal/km2/day * 2500 km2 * 0.1 (loss from 1 trophic  level)                                       * 0.001 (loss from solar absorption percentage) ]
Mmax = (780812500 Cal/day) / (1000 Cal/day)

Mmax = 780 812 people