Bales Needed
1,242
Tons
31.1
Barn Length
24 ft
Feed Days
180
Bales Needed
1,242
31.1 tons total
Daily Feed
300 lbs
Feed Days
180
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Inputs
Result
Daily = 30 × 10 = 300 lbs. Total = 300 × 180 × 1.15 = 62,100 lbs. Bales = ceil(62,100 / 50) = 1,242.
Inputs
Result
Daily = 4 × 6 = 24 lbs. Total = 24 × 150 × 1.15 = 4,140 lbs. Bales = ceil(4,140 / 50) = 83.
A 1,200 lb cow eats about 30 lbs of hay per day. Over a 6-month (180-day) winter, that is 5,400 lbs per cow. With 15% waste, you need about 6,210 lbs or 124 small square bales (50 lbs each) per cow.
| Animal | Daily Hay | 6-Month Bales (50 lb) | 6-Month Tons |
|---|---|---|---|
| Cattle | 30 lbs | 124 | 2.7 |
| Horse | 20 lbs | 83 | 1.8 |
| Goat | 4 lbs | 17 | 0.4 |
| Sheep | 4.5 lbs | 19 | 0.4 |
Barn size depends on bale type and count. Small square bales stack efficiently at 3 ft long by 1.5 ft wide by 14 inches tall. For 500 bales stacked 10 high in a 24 ft wide barn, you need about 25 ft of length plus aisles.
Hay waste varies by feeder type. Open ground feeding wastes 30–40%. Ring feeders waste 10–15%. Cone feeders or hay nets waste 5–10%. Budget at least 15% waste for typical ring-feeder operations.
Small squares (50 lbs) are easier to handle by hand and stack tighter with less waste. Round bales (1,000 lbs) are cheaper per ton and faster to move with equipment. For fewer than 10 cattle, small squares are usually more practical.
| Bale Type | Weight | Cost/Ton | Equipment Needed |
|---|---|---|---|
| Small Square | 50 lbs | $200–320 | None (hand stack) |
| Large Square | 800 lbs | $150–225 | Tractor + forks |
| Round | 1,000 lbs | $80–160 | Tractor + spear |
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Last Updated: Mar 21, 2026
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