Hay Production

Farmers love to make hay but they seldom forget that winter feeding is the highest cost in their operation.  Hay is often cheaper than alternative feeds especially if it meets the animal requirements without additional supplementation.  There are management practices that can limit the amount of hay needed, but it is still essential if only to avoid disaster for those stockpiling tall fescue or using Rotational Grazing. 

For cool season grasses, the spring cutting is normally the highest yielding, but fall cuttings can be equal in tonnage when fertilized.  Water is what typically limits fall production.  First cutting hay is usually done when the first seed heads emerge, for tall fescue in our area, that is around May 15th, but variable.  When hay is cut in that stage or earlier, it will be high quality with more nutrient value.  In terms of maturity orchard grass is earlier than tall fescue which is earlier than timothy.  In the fall, fescue keeps its nutrient value the longest, while timothy's yield is very low compared to its spring yield in this climate. 

One of the best tips for hay production is not to cut the stand too low.  Low cutting will prolong recovery and can thin the grass stand.  A height of 4" is often thought sufficient for the health of the plant for most grasses. 

So, How Much Hay?

So what we would like to do here is approach things cost effectively, so that hay fields get fertilized appropriately.  We can use formulas to estimate yield and determine the proper fertilization.  Producers will need to answer the question, "How much hay do you need?".  After that everything becomes a calculation and will work itself out. 

If you already know how much hay you need, you are all set but if you are uncertain, a formula can help determine how much and convert it to dry matter tons.  A few useful assumptions for round bales: 5x4 = 880lb.; 5x5 = 1100lb.;  5x6 = 1600lb.  So if you were trying to determine how much you needed from how much your herd will eat this winter, another assumption is needed, and that for this example we will use cattle and we will assume that cattle eat 2.5% of their body weight per day, for other types of livestock please use the necessary intake coefficient in % body weight.

If you had a herd of 25 cows, estimated to average 1200lb, and you estimate that you will have to feed 5 months or 150 days.  The assumption is based on no pasture production or significant supplementation during the 150 day period, so please adjust as necessary.  The amount of hay in tons will follow this equation below:

Tons of Hay Needed = number of cows x cow average body weight x .025 x number of feeding days / 2000 lb.

So from the beef cow example: 25x1200x.025x150/2000 = 56.25 tons

To convert that to the number of bales needed, say you used 5x5 round bales: just divide 2000 lb. by 1100 lb. and multiply by 56.25 = 102.3 bales

If you already knew you needed, say 103 bales; to convert the number into tons, just multiply 103 by 1100 lb. then divided by 2000 lb. = 56.65 tons

The reason why everything is converted to tons is because fertilizer recommendations are published based on tons of dry matter.  So to determine how much fertilizer you should put down, we calculate how much hay we need per acre in dry matter tons, and multiply it by the nutrient removal rate.  A few necessary assumptions about the amount of plant nutrients removed per ton of hay in N-P-K ratios: Tall Fescue 39-19-53, Orchard Grass 50-17-62, Bermuda 43-10-48, Alfalfa 56-15-60 (note Alfalfa provides most of its own nitrogen while the others do not).  Using the formula below and for an example we need 56.25 tons of tall fescue from 25 acres of hay field.

Desired Hay Tonnage per Acre = Number of Hay Tons Needed / The Number of Hay Acres Grown

In our example: 56.25/25 = 2.25 tons per acre

So to convert that into fertilizer per acre we just multiply the required yield in tons per acre by the assumed removal rate.  So from the tall fescue example multiply each of the N-P-K values by the desired tons per acres.

In our example: 2.25 x 39 - 2.25 x 19 - 2.25 x 53 = 88N - 43P - 120K

In the example it would be common for a producer to fertilize in the spring with 70N - 35P - 105K and hope to make most of the tonnage and without fertilizing in the fall, get the desired total tonnage.  Other producers prefer to put down P&K to winterize the grass in the fall and apply only nitrogen in the spring.  No matter what the practice it takes about 2 years to see the big effects on the grass stand from using P&K. 

The yield you get will not necessarily be what you expect.  Rain, soil temperature, and other things will effect it.  If you yield more than expected, your soil nutrient levels will fall, if you yield less, they will rise.  Nitrogen is the only product that is completely lost no matter what the yield.  That is also why it is considered the most limiting of the 3 major plant nutrients.

Soil nutrient levels will also affect how the plants respond to fertilizer.  Soils at medium levels will have a much better response than soils at low nutrient levels.  Please see Soils.  When the fertilizer applied is the same, higher yields will result from higher soil nutrient levels because more fertilizer units will be readily available rather than tied up in chemical reactions with the soil.  In the same fashion and even more pronounced, the pH will affect the availability of fertilizer units.  For grasses the ideal pH is around 6.1 - 6.3.  For legumes, like alfalfa, the ideal pH should be around 6.3 - 6.5.  For more information please see Bulk Lime.