June 13, 2013
First cuttings of alfalfa are later than normal in many areas. When this happens, alfalfa weevil activity often increases.
Figure 1. Alfalfa weevil adult |
Weevil larvae generally eat small holes in leaves at the growing tip of alfalfa plants. As larvae grow, damage increases as the holes become larger. Severely damaged fields have a grayish appearance because of the drying of skeletonized upper leaves and buds.
Most fields did not have enough damage to need spraying before first cutting. After harvest, many larvae die when exposed to direct sunlight and high temperatures at the soil surface. But will enough larvae survive to be a problem for regrowth?
A greater concern may be the survival of weevil adults. These critters can be really hard on alfalfa regrowth by feeding on the developing crown buds, retarding growth and preventing fields from greening up after harvest.
The first step in controlling weevil adults is the removal of windrows and bales as soon as possible to expose these insects. Examine stubble frequently to see if adults are delaying green-up by feeding on new buds, especially where the windrow laid.
If there is damage, use Table 1 to help determine when spraying may be desirable. This guide considers insecticide cost, hay value, and harvest management.
Don’t let alfalfa weevil adults delay second growth of your alfalfa. Keep a watchful eye on the stubble and spray, but only if necessary.
For more information see Managing the Alfalfa Weevil (G1208).
Bruce Anderson
Extension Forage Specialist
Table I. Alfalfa stubble threshold calculation chart | ||
Factors | Example | Your field |
A. Insecticide plus application cost ($/acre) | $7 | ________________ |
B. Value of hay ($/ton) | $100 | ________________ |
C. Loss factor (1st bloom harvest = 0.0198; 28-day harvest = 0.0345) | 0.0198 | ________________ |
D. Days of complete defoliation that can be tolerated* | 3.5 | ________________ |
*To estimate D, multiply B times C and divide into A. Using the numbers in the example column, the calculation for the number of days that complete defoliation that can be tolerated is: D = A/(BxC) = 7/(100 x 0.0198) = 7/1.98 = 3.5 days |