A Simpler Approach
(Lesson 5. Grazier’s Arithmetic continued)
For those who find equations a bit intimidating or take a very simple approach to life, forage allocation can be made simpler. If we compute the average for each stand density from Table 1, we come up with approximately 175, 275, and 350 lbs./acre-inch for fair, good, and excellent pasture, respectively. If we assume that a 1000 lb. lactating cow will consume around 3% of her bodyweight, we can figure a “cow-day” to be equal to 30 lbs. of forage consumed (see Table 2 for livestock consumption data). If we divide 30 lbs. of forage/cow-day into the lbs. of forage /acre-inch, we find that the “cow-day” yield of fair, good, and excellent pasture to be about 6, 9, and 11 cow-days/inch of pasture consumed.
This becomes a simple method of allocating pasture:
- Look at the pasture and determine it to be fair, good, or excellent.
- Measure or estimate the height of the pasture to be allocated.
- Subtract from the total height the height of stubble you want the animals to leave.
- Multiply the difference between starting height and ending height by the cow-days/inch to figure available cow-days/acre.
- Divide the number of cows in the herd by cow-days/acre to figure how much area should be allocated.
EXAMPLE:
Step 1. We look at the grass and say, “This is average (good) grass” which gives us a cow-day/inch factor of 9.
Step 2. We measure the height to be 8″.
Step 3. We would like to leave a 3″ residual, so 8″ – 3″ = 5″ to be grazed.
Step 4. 5 inch grazed X 9 cow-days/acre-inch = 45 cow-days/acre.
Step 5. If we have 100 cows, we should allocate (100 cows/45 cow-days/acre)= 2.2 acres/day.
Table 2. Livestock consumption data*
(pounds daily dry matter intake based on body weight and level of performance)
| Species | 3.5% | 3.0% | 2.5% |
|---|---|---|---|
| Dairy cow (1400 lbs) | 49 | 42 | 35 |
| Dairy heifer (1000 lbs) | 35 | 30 | 25 |
| Sheep (250 lbs) | 8.75 | 7.5 | 6.25 |
| Sheep (125 lbs) | 4.4 | 3.75 | 3.1 |
| Beef cow-calf pair (1600 lbs) | 56 | 48 | 40 |
| Beef cow-calf pair (1200 lbs) | 42 | 36 | 30 |
| Beef steer (600 lbs) | 21 | 18 | 15 |
| Horse & equines (1000 lbs) | 35 | 30 | 25 |
| Horse & equines (1600 lbs) | 56 | 48 | 40 |
| Goat (75 lbs) | 2.6 | 2.25 | 2.0 |
| Goat (150 lbs) | 5.25 | 4.5 | 3.75 |
| Elk (800 lbs) | 28 | 24 | 20 |
| Red Deer (220 lbs)? | 7.7 | 6.6 | 5.5 |
| Fallow Deer (150 lbs) | 5.25 | 4.5 | 3.75 |
| Alpaca (250 lbs) | 8.75 | 7.5 | 6.25 |
| Llama (500 lbs) | 17.5 | 15 | 12.5 |
* For accuracy check the weights of your own animals.
If you are going to use a flexible paddock system where temporary fences are strung between two permanent subdivision fences, it is very desirable to set the main fences at a spacing that allows for simple calculation of acre increments. It is very handy to have the line posts in the permanent fences set at known intervals that mark fractional acre increments. For example, if the permanent strips are 300 feet wide, then line posts at 50 foot intervals mark about 1/3 acre strips. Forward planning at fence building time can make future operation of the system much simpler.
This example works fine for 1000 lb. cows, but what about other classes of livestock. As grazing management is an imprecise science due to ever changing conditions, we are only looking for an approximation, not perfection. If cows are anywhere in the 900 to 1200 lb. range, the simple approach used above will probably be adequate. If you have livestock of different weights, you can add up the total estimated weight of all the livestock in the herd, divide by 1000 and be fairly close to “cow-day” equivalents.