Showing posts with label Calories. Show all posts
Showing posts with label Calories. Show all posts

05 August, 2009

Food Foot Print

Assuming that the diet is balanced, yields are average and land is 1k, the calculation of the area of arable land needed per person is as follows:


Item Grams/Day Kgs/Year Yield Land

Wheat 100.00 40.00 400 400
Rice 62.50 25.00 1000 100
Millet 37.50 15.00 600 100
Barley 75.00 30.00 600 200
Veg 62.50 25.00 1000 100
Oil 31.25 12.50 250 200
Sugar 31.25 12.50 500 100
Spices 31.25 12.50 250 200
Fruits 250.00 100.00 667 600

SUM 2000


Notes:
(i) The figures for land is given in sq meters, there are about 4000 sq meters in one acre so the above sum is half acre.

(ii) The sum for all grains is 800 sq m (400 for wheat, 200 for barley, 100 each for rice and millet) but we should take 900 sq m because a substantial amount of grains grown goes in shell, for eg the shell of wheat removed in milling.

(iii) The grains includes barley and millet because they are nitrogen fixing crops.

(iv) Oil is vegetable oil. Since vegetable yield is 1000 kg and oil extraction rate is 25% so 250 kg is its yield.

(v) The above calculations are based on a 400 days diet per year, the extra number of days allow more-than-normal eating in feasts and parties, roughly 10% of the entire year.

(vi) For simplicity straight forward figures about crops directly eaten by humans is given first, the more complex subject of crops grown by humans but eaten by animals and then those animals and their milk consumed by humans is discussed below.

It should be known that per kg mass animals and infact insects and birds too require same number of calories as humans for same level of activities. So a cow which is lactating needs 60 calories per kg per day and which is not lactating needs 40, so average is 50. It should also be noted that for each kg gain in both animals and humans same 7,500 calories are needed. Since about 60% mass of an animals is meat (including bones), therefore 12,500 calories are needed per kg meat at minimum.

Another important point to consider is calories in hay, but first what is hay? Hay is a general term that includes all grass, straw, leaves etc eaten by animals but not all grass, straw and leaves are hay. Fresh grass which have visible moisture on its surface is only 15% dry matter and 85% moisture, fresh grass and leaves with no visible moisture contains 20% to 30% dry matter and rest is moisture, brown grass and leaves contains may be 50% dry matter. The point is, when fresh moisture level is high and when time pass and the green matter becomes brown its moisture level decreases. It should be kept in mind that the energy needed by the animals is only in the dry matter, there is no energy in moisture.

Hay is grass, straw or leaves that has atleast 80% dry matter but dry matter may rise upto 90%. For our calculations we ignore the higher proportion and assume that only 80% of hay by mass is dry matter to be safe. Each gram of dry matter contains 2.5 calories. So the hay as a whole contains 2 calories per gram.

It should be noted that all over this website the term "calorie" actually refers to "kilo calorie" which is equal to 4200 joules and has enough heat energy to raise temperature of 1 kg of water 1 degree celsius.

When grain is grown along with it an equal amount of straw is grown. This is the covering substance in which the grain is contained. Note that it is different than the shell of the grains which is still present in the cultivated crop. For example if you buy wheat (not flour) from market the covering shell is still present but the straw was removed at the time of cultivation. The straw is already hay, that is, it already contains 80% by mass dry matter and each gram of it has 2 calories.

For all grass and grass like crops grown with prominent exception of alfalfa, the same 1600 kg hay is grown per acre per year at 1k land. Note that we only are concerned with the mass of hay grown, not the mass of fresh grass etc. The trick is to store it until it lose all the moisture it can, then what is left is hay. It means that usually not more than 10 to 20 percent of moisture is left in grass and grass-like crops and not all the moisture is lost in storage.

In case of leaves, the hay yield is 600 kg per acre per year provided the land is properly covered with trees. By properly I mean the usual number of trees per acre as used to be the norm in traditional (pre-1950) agriculture. In an orchard and also in a forest, that is, in all tree lands, another 200 kg grass hay is also grown per acre per year. Note that the figure 200 came from the natural figure of 400 kg hay yield in natural pastures halved. It is halved because due to the presence of trees only half as much grass can be grown as would be in a tree less pasture. Natural pasture is a tree-less land where grass and grass-like crops are naturally grown without any farming by humans. Note that when humans do farming in a traditional way the yield is already 4 times than what is grown naturally.

Last thing to consider is that after complex calculations and data reviews I came to the conclusion that 2.56 kg hay is the minimum amount of hay needed to produce 1 kg cow milk assuming there is no meat production, but since meat production cannot be avoided as some of the food eaten by cow converts into a calf and the cow itself needed food to grow up when it was a calf and not producing any milk therefore a relation of 3.2 is the minimum practical ratio, meaning 3.2 kg hay results in 1 kg milk and along with it 1/16 kg meat. Note that in case of cows our emphasis is on milk production, not meat production. Also note that since the "meat" produced in this case would not be pure meat but would be containing bones too which are not edible by humans and since in our calculation above we only consider pure meat therefore we actually have to produce more meat-with-bone to have the desired amount of pure meat. In simple words we have to increase the ratio to 3.6. Note that if we make it 4 then milk production would be same but meat production would be 1/8 of milk production.

We want to produce 200 kg milk, half of which that is 100 kg would be drunk directly and rest half would be converted into ghee and butter. To produce 200 kg milk we need 200 x 3.6 = 720 kg hay. In above calculation we account for 110 kg grain production which means 110 kg straw-hay is produced. The straw is supposed to be stored for maximum 6 months so that food is available for our cows in autumn and winter when fresh grass-hay is not available. During storage at maximum 20% of the straw may lose its nutrients therefore we take 80% of that, 0.8 x 110 = 88 kg straw. So, we have 88 kg straw available for our cows.

We keep some land in form of pasture but that is not natural pasture. It is called pasture only because certain grass like crops would be grown on it for our cows. We keep a 1600 sq m patch of land for this purpose. Since per acre yield of grass hay is 1600 kg, therefore 1600 sq m or 0.4 acres produce 640 kg grass-hay.

We add the straw-hay and grass-hay, 88 kg + 640 kg = 728 kg. It is enough to produce 200 kg cow milk and along with it beef and beef-fats. The ratio between milk and hay is 3.6, so 200 kg x 3.6 = 720 kg. Note that about 89% of the cow food is grass or grass-like crops so unlike western industrial farming we don't feed grains to the cows which their stomach are not made for.

We also have an orchard whose size is 600 sq m. Since the per acre yield of orchard hay is 800 kg, from 600 sq m we can get 120 kg hay. Note that this is fed only to goats to have mutton and mutton-fats. Along with that there is ofcourse some opportunity to get goat-milk but we ignore that to have some flexibility. It is known that each kg of edible beef (with bones) requires atleast 12 and usually 16 kg hay to be produced and each kg of edible mutton requires 10 kg hay usually. So 120 kg hay results in 12 kg mutton. I take that as 12.5 kg mutton to get the daily figure of 31.25 grams.

While we are on this it is suitable to calculate the other crops needed per capita which are not included in food items. They are tea/coffee and cotton. Tea/coffee is considered a non-food item because it contains no calories at all. The world per person average of cotton consumption in year 2000 was 3.125 kg per year. Ofcourse non-cotton clothes items are also consumed by people included silk etc but for the purpose of this calculation we ignore all those and assume that only cotton is consumed, because we are ignoring all other clothes items we double the per capita cotton consumption estimate to 6.25 kg. In other words, it can be said that cotton is consumed as usual and an equal amount of other clothes items are also consumed. Per capita tea/coffee consumption figure is taken as 12.5 kg per year which is closed to reality.


Item Kgs/Day Yield Land

Cotton 6.25 250 100
Tea 12.50 500 100


So in addition of the 2000 sq m land needed per person shown in chart above, a 1600 sq m pasture and 100 sq m tea farm and 100 sq m cotton farm is also needed, this brings the figure to 3900 sq m. To be one acre a 100 sq m is still lacking which I leave to readers to fill in an item of their choice. It could for example be grain farm to support chicken which should result in 12.5 kg chicken meat or a pond of fresh water fish which should result in 12.5 kg fish meat or dry fruits farm which should result in 12.5 kg dry fruits.

In short, to support one person, we need on average, 1 acre of arable land, that is of category 1k meaning it either has 10 inches rain or 1 acre-ft canal water or a combination of them providing 800 cubic meter water to crops.

I stress that this is the natural arable land need per person. We can see from history that in case of subcontinent (land comprising today's india, pakistan, bangladesh, afghanistan, burma and nepal) in mughal era (1526-1739), that is, when the entire land was ruled by mughals, a population of 100 million was supported by 400 million arable acres. This give us the figure of 4 arable acres per person, how was it? Well, since by large due to absence of refrigeration a large amount of food cannot be stored so 2 arable acres were kept per person in a village, that was the norm. 2 acres so that if crop fails and half crop is grown then too the village population could be supported, usually in case of crop failure over a wide area of land the yield falls to half, not to quarter or zero. It is not to deny that locally that is in a village almost no crop is grown but due to presence of trade the risk is shared over a district so as a whole half crop is failed and half is saved. It can be seen in another way too that for example grain crops failed but the fruits and vegetables and animal products were still there. So, 2 arable acres per person in a village. In case of usual crop, twice-than-village-need crop is grown which is exported to cities. The other 2 arable acres were left for wild life, in form of forests, natural pastures, un-touched lakes etc.

28 July, 2009

Yields

The following data is based on 1k Land. It is land that gets either 10 inches of rain water 80% of which falls in crop season or 1 acre-ft/acre water 33% of which may evaporate in transit, effective water that is actually used by land is therefore 800 cubic meters per acre per year. The land gets no ground water. There is no spreading of artificial fertilizers. No pesticides are used. The seeds are traditional/pre-green-revolution. Only one crop per year happen, planted in spring and cultivated in autumn.


Yields Water Need Total Water Need
(kg/acre/yr) (cu-meter/kg) (cu-meter/yr)
X Y XY

GRAINS:
Wheat 400 1.00 400
Rice 1000 5.00 5000
Millet 600 0.70 420
Barley 600 0.70 420
Pulses 250 1.40 350

Veg-Oil 250 2.80 700
Spices 250 2.80 700
DryFruits 250 2.80 700
Sugar 500 1.40 700

Tea/Coffee 500 1.40 700
Cotton 250 10.00 2500

Grass-Hay 1500 0.70 1050

Fruits 1000 0.70 700

Veg 1000 0.70 700

Notes:

(i)In some crops not all of the effective water is used. Such crops includes wheat, millet, barley, pulses etc, the so-called dry crops. Due to their less use of water these crops can sometimes be planted in dry weather of winter where rainfall is less than summer. These crops can also be planted in poor land having only 400 cubic meters water. It not means that the yield would be same in poor quality lands. The traditional seeds of dry crops unlike the green-revolution seeds are not capable of using all of the available water.

(ii)In case of fruits 700 cubic meters water is used out of the available 800 cubic meters, the rest is used by grass and weeds that grows in places between trees.

(iii)In case of grass more than 800 cubic meters of effective water is used. Due to the typical net-like pattern made by grass roots underground water absorption in ground is reduced.

27 July, 2009

Types of Land


We can divide land on the basis of the amount of water it receive. The underlying assumption is that upto a certain extent fertility of land grows with amount of water available to it. Desert having little water has little fertility whereas deltas of nile, euphrates, ganges and indus have high fertility due to water availability.

There are infact three sources of water, primary which is rain, secondary which is canal, teritiary which is ground water. I would not consider the teritiary source in this article for land classification because it is energy intensive.

On the basis of quantity of rain it gets a land can be divided into 7 categories:

>80 inches
as good as desert for food production

80 inches
not so fertile due to excess water, can only grow a few crops like rice, grass, poor quality

40 inches
average land

20 inches
the most fertile land

10 inches
average land

5 inches
poor quality land

<5 inches
desert

On the basis of canal water available the land can be divided into 3 categories:

0 acre-ft/acre

1 acre-ft/acre

2 acre-ft/acre

What is "inches of rain" and "acre-ft/acre" anyways?

Lets consider the middle case of 10 inches of rain falling on one acre. Since there are 2.54 cm in each inch therefore it means 25.4 cm tall water column standing on one acre. Lets round that off to 25 cm for easy calculation and understanding. One acre is 4840 sq yards which is equal to 4047 sq m. Lets round that off too to 4000 sq meters. So, 10 inches of rain roughly means 25 cm water standing on 4000 sq m. The volume is 1000 cubic meter water. I call this 1k land because it gets 1000 cubic meter water in a year. In my calculations 1k land can be any land that gets 1000 cubic meter water irrespective of the source of water, it not always have to be rain water to call it "1k land".

One acre-ft/acre canal water means 1 ft tall water column standing on 1 acre of land. Acre-ft is the general unit of measurement of canal water. 1 ft means 12 inches and one inch means 2.54 cm, so 1 ft means 30.48 cm. Round that to 30 cm. Multiply that with 4000 sq m as before and we get 1200 cubic meter water per acre.

Before ending this discussion lets find the cost of rounding. In case of rain water, if we multiply the actual 4047 sq m with the actual 25.4 cm we gets 1028 cubic meter water instead of the round off value of 1000 cubic meter. In case of canal water, if we multiply the actual 4047 sq m with the actual 30.48 cm we gets 1234 cubic meter water instead of 1200 cubic meter.

Effective Water

In addition of knowing how much water is available per acre one must also know how much water getting to land is actually available for crops. Ofcourse not all of it is, some rain fall in off season when its of no use to crops, water flowing in canals and before that in rivers get evaporated on their long (hundreds of miles) journey to farm etc. Lets calculate the effective water in both cases of rain water and canal water one by one.

In my part of world, that is, in south asia, 80% of all rain water falls in two months of sawan and badho, that is from 16th july to 15th september. This also used to be the time when the crops need water in the traditional one-crop-per-year farming. So, 80% of all rains gets to land when the crops need them. I assume for simplicity the same for the rest of the world. It means that the average case of 10 inches rain water per year which is roughly translated to 1000 cubic meters water per acre per year is 800 cubic meters effective water.

In case of canal water, in Punjab due to relatively cooler weather and more fertile land the evaporation rate of canal water is 25%, in Sindh due to relatively hotter weather and generally less fertile land and also due to longer distance the canal water has to travel to farm the evaporation rate of canal water is 33%. I, for simplicity, takes the general rate of evaporation 33% so that 1 acre-ft/acre water which is roughly translated into 1200 cubic meters water above means 800 cubic meters effective water to equate with 10 inches rain. Note that there is no off-season canal water distribution because canal water is essentially the rain water stored in dams and barrages distributed only when needed, so all of it gets to land when needed there is no discount of off-season water, the only discount is evaporation.

Lets go back to the discussion of categories of land. Excluding desert and desert-like lands where rain water is less than 5 inches or more than 80 inches, we get five categories of land on basis of rain water. We also have 3 categories of land on basis of canal water. So, taking union, we have 15 categories of land as follows:


5 inches rain = 400 cu m effective
5 inches rain + 1 acre-ft canal = 1200 ditto
5 inches rain + 2 acre-ft canal = 2000 ditto

10 inches rain = 800 ditto
10 inches rain + 1 acre-ft canal = 1600 ditto
10 inches rain + 2 acre-ft canal = 2400 ditto

20 inches rain = 1600 ditto
20 inches rain + 1 acre-ft canal = 2400 ditto
20 inches rain + 2 acre-ft canal = 3200 ditto

40 inches rain = 3200 ditto
40 inches rain + 1 acre-ft canal = 4000 ditto
40 inches rain + 2 acre-ft canal = 4800 ditto

80 inches rain = 6400 ditto
80 inches rain + 1 acre-ft canal = 7200 ditto
80 inches rain + 2 acre-ft canal = 8000 ditto


So, in short, the amount of effective water available vary between 400 cu m to 8000 cu m, a factor of 20. What we need is not the average case but the typical case. The average case is useful when the good, fair and worst are all in the same proportion which is not a real life situation.

So what is a typical case? A typical case is the mode as in statistics, that is the most common situation. It not need be the mathematical average. Its the case which you are most likely to encounter in a given situation.

How to find the typical case in a exponential situation? Well, first what is the exponential situation? Exponential situation is the situation where the numbers vary in multiples, like a sequence of 1,2,4,8,16. Here the typical case is 4 which is the middle value. It is the situation which you are most likely to encounter.

In the discussion of category of land, the best approach is to combine the case of 10inches of rain with 1 acre-ft/acre canal water. Result is 1600 cu m effective water per acre per year. For a deeper discussion, in order to get a sustainable agriculture it is unwise to use the canal water on regular basis. The agriculture should be primarily based on only the rain water, keeping the canal water as a safety valve in case of emergencies. Emergencies in this case includes droughts, pest attacks etc. If we embed the safety valve, the reserve troops in the normal operations where would we fall in case of failures? It sure is inefficient but it is indeed very resilient. All the world agriculture before the 1950s was infact based on only rain water, canal water was used only in emergencies, in years when rainfall is less than normal. Other disadvantages of canal water would be discussed in a separate article.

In summary, in absence of canal water, the average effective water available is 800 cu m, the lower value is 400 and the higher value is 1600, so we get the simple pattern of 3 values.


Worst: 5 inches OR 0.5 acre-ft = 400 cu m eff

Typical: 10 inches OR 1.0 acre-ft = 800 ditto

Best: 20 inches OR 2.0 acre-ft = 1600 ditto
OR
10 inches + 1.0 acre-ft = 1600 ditto

Average: 40 inches OR 4.0 acre-ft = 3200 ditto
OR
20 inches + 2.0 acre-ft = 3200 ditto

Worst: 80 inches OR 8.0 acre-ft = 6400 ditto
40 inches + 4.0 acre-ft = 6400 cu m ditto


Note that 3200 cu m case is taken as average because after a certain level excess water actually reduce fertility instead of increasing it.

Only a few of the possible combinations are given above for 3200 cu m and 6400 cu m.

The point is hidden in the amount of effective water available, not on the source of water.

For simplicity, some new terms are introduced, 2k means 2000 cu m nominal and 1600 cu m effective, 3k means 3000 cu m nominal and 2400 cu m effective and so on. In that sense, land types varies between 1k and 10k, 2k being most fertile, both 1k and 4k being average, both 8k and 0.5k being poor. More than 8k are extremely unlikely to be of any value to agriculture.

Another important thing to consider is the relation between intensity of rain water and availability of canal water. At fertile land, for example in punjab, both the availability of canal water and intensity of rain is higher than that in sindh. An increased amount of canal water applied to land increase the humidity of air which inturn results in more rain fall. Considering the feedback loops would increase complexity so are not discussed in detail in this article. Only take-home message from this is that it is more likely to get 2 canal water in a land that already has 20 inches rain than in land that has 5 inches rain because that would be a desert already having little or no access to any kind of river or canal.

24 July, 2009

Calories Part 2 - Diet Plan


Following is a daily balanced diet plan for an average 2000 Calories food requirements:



ITEM GRAMS CH P F Calories

Wheat 100.00 66.67 12.10 2.70 339.92
Rice 62.50 43.75 6.25 2.72 225.02
Millet 37.50 8.88 1.32 0.37 44.20
Barley 75.00 55.12 9.38 1.72 273.82

Milk 250.00 12.20 8.30 9.10 165.72
Butter 5.12 - - 4.10 37.72
Ghee* 4.55 - - 4.10 37.72

Mutton 31.25 - 8.10 0.27 34.88
Beef 31.25 - 25.00 - 100.00

Fats** 7.81 - - 6.25 57.50

Veg 62.50 3.91 - - 15.63
Veg Oil 31.25 - - 31.25 287.50
Sugar 31.25 31.25 - - 125.00
Spices 31.25 15.63 6.25 - 87.50

Fruits 250.00 31.25 - - 125.00

SUM 971.50 268.66 76.70 62.58 1957.15




*Ghee is fats taken out from milk. In the above diet plan 125 grams of milk is dedicated to make ghee. 125 grams milk contains 4.55 grams fats but only 90% of them is taken out, rest is lost in the process. Ghee itself contains 90% fats, rest is water. Therefore 4.55 grams fats in original milk results in 4.55 grams ghee. In the same way 125 grams milk is dedicated to make butter, the 4.55 grams fats present results in 4.095 grams fats in butter after 10% loss of fats. Butter is 80% fats so 5.12 grams butter.


**Animal fats is the fats available in the fats in the skins and other organs of animals. They make up 20% of the body mass of animals. Assuming that it is not desirable to use all of these fats we go for only 7.5%. Since animals are by mass 60%meat therefore animal fats are 1/8 of the meat.

Note:

The figures may look odd for daily requirements but if multiply by 8 they become more easy to comprehend. For eg vegetable oil 31.25 grams daily means 250 grams per 8days. The figure 8 is to roughly calculate for a week 10 being so far from 7. It is assumed that the food buying is done once per week. The oddness of figures should also get removed in next article about land requirements for growing the stated food items, the figures being tuned to yields.

Calories Part 1 - Basics


There is a unit of energy called "calorie" which is equal to enough heat energy needed to raise temperature of 1 gram of water 1 degree centigrade. In diet calculations a higher unit of energy "Calorie" (with capital C) is used which is equal to 1000 calories, enough heat energy to raise temperature of 1 kg water 1 degree centigrade.

In diet requirements, an organism needs 30 to 50 Calories per kg body mass. This is true for all animals and humans. A person doing light work such as table work needs just 30 Calories whereas a farmer or a carpenter needs 50 Calories. For our calculations we take the average 50 Calories.

Other than noting the overall energy need its also important to know the components of food needed. There are seven components of food: Carbohydrates, Proteins, Fats, Minerals, Vitamins, Fiber and Water. The last two, Fiber and Water, contains no energy at all, fiber is needed to rest the system as system not have to digest the fiber, water is needed to keep the digestive system clean. Minerals and Vitamins also contains no energy but are needed to rebuild the body after daily wear and tear. The only energy contents are first three, in this article we discuss only them.

Each gram of Carbohydrate contains 4 Calories, each gram of Protein also 4 Calories, each gram of Fats 9.2 Calories. For a balanced diet, 56-60 percent of energy must come from Carbohydrates, 12-15 percent from Proteins and the rest 25-32 percent from Fats.

The amount of diet energy needed depends of level of activity which is constant for a given person. A farmer remains to be a farmer and a clerk clerk. The other thing energy depends on is body mass which in turn depends on height which in turn depends on age if the body is growing.

Lets take the simple case of an adult who have reached the maximum height it would acquire in his/her lifetime. This adult would have in most cases also fixed its activity level by choosing a profession and gain enough mass appropriate to height. In most cases the optimum mass of 5'8" to 6' man is 75 kg and a naturally shorter woman of equal age 62.5 kg. For the population as a whole the average ideal mass is 50 kg, that is, if you consider children too.

Lets take 50 kg average mass and average activity level, that translates into a 2000 Calories daily energy need. Note that even for a 75 kg adult mass if doing office work the energy requirement is close to 2000 Calories, actually 2250 Calories. For a 2000 Calories, 1200 Calories must come from Carbohydrates (60%) which means 300 grams of it and 300 Calories from Proteins (15%) which means 75 grams and 500 Calories from Fats (25%) which means 55 grams.

So, in short, for an average person, the need is 300 grams Carbohydrates, 75 grams Proteins and 55 grams Fats. Each of these components of food available in different food items is given below in percentages:



ITEM CH P F Calories

Honey 80.00 - - 3.20
Milk 4.88 3.32 3.64 0.66
Wheat 66.67 12.10 2.70 3.40
Rice 70.00 10.00 4.35 3.60
Millet 23.68 3.51 0.98 1.18
Barley 73.48 12.50 2.28 3.65
Pulses 50.00 26.00 6.00 3.60
Fruits* 12.50 - - 0.50
Others 25.00 - - 1.00
Veg 6.25 - - 0.25
Mutton - 26.00 0.87 1.12
Beef - 80.00 - 3.20
Fish - 18.00 - 0.72
Butter - - 80.00 7.29
Sugar 100.00 - - 4.00
Oil - - 100.00 9.20
Spices 50.00 20.00 - 2.80



*This is an average calculation for different types of fruits not including those having a higher sugar amount such as banana/mango etc

Another thing to consider in getting the food is how much load it put on soil, in short how much land is needed to support an average person.