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Balancing Horse Diets

Discover how to balance your horse's diet for proper digestive function, preventing colic, laminitis, and behavioral issues. Learn about feeding guidelines for various horse categories and the importance of evaluating nutrients and energy values of feedstuffs.

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Balancing Horse Diets

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  1. Balancing Horse Diets • All horses should be fed a minimum of 1% BW of forage per day • Proper digestive function • Prevents colic, laminitis • Prevents behavioral problems

  2. Balancing Horse Diets • Horses at maintenance: • Will eat approx. 2-3% BW per day • May get all required Lysine, CP, and energy from forage

  3. Balancing Horse Diets • Some horses may need added grain and protein/Lysine • Lactating mares • Growing horses • High exercise demands • “Poor keepers”

  4. Balancing Horse Diets • Fat can be used to add energy • Replaces starch • Helps prevent colic and/or founder • Max amount of fat = 10% of diet DM

  5. Balancing Horse Diets • Once we’ve provided 1% BW forage and suggested amount of fat • Evaluate nutrients supplied • Compare to requirements • Balance the rest of the diet

  6. Balancing Horse Diets • Mature 500 kg gelding at moderate work • Requirements: • 24.6 Mcal DE • 984 g CP

  7. Balancing Horse Diets • Feedstuffs: • Grass hay (1.95 Mcal/kg DE, 8.46% CP) • Alfalfa hay (1.75 Mcal/kg DE, 18% CP) • Grain (2.45 Mcal/kg DE, 8.9% CP) • Vegetable oil (9.2 Mcal/kg DE)

  8. Balancing Horse Diets • Energy Values of Feedstuffs: • Use energy values specifically for horses • Lower energy values than ruminants • TDN, % ruminant x 0.88 = TDN, % horse

  9. Balancing Horse Diets • Feed 1% BW as grass hay • 500 kg x 0.01 = 5 kg grass hay intake • 5 kg x 1.95 Mcal/kg = 9.75 Mcal DE • 5 kg x 0.0846 x 1,000 g/kg = 423 g CP • Add 0.5 kg vegetable oil • 0.5 kg x 9.2 Mcal/kg = 4.6 Mcal DE

  10. Balancing Horse Diets • Need 24.6 Mcal – 9.75 Mcal – 4.6 Mcal = 10.25 Mcal DE • Need 984 g – 423 g = 561 g CP

  11. Balancing Horse Diets • A = kg alfalfa hay, B = kg grain • 10.25 Mcal = 1.75 Mcal/kg A + 2.45 Mcal/kg B • 0.561 kg = 0.18 A + 0.089 B • 1.75/0.18 = 9.72 multiplication factor for equation 2

  12. Simultaneous Equations • 9.72 (0.18 A + 0.089 B) = 9.72 (0.561 kg) • 1.75A + 0.865B = 5.45 kg new equation

  13. Simultaneous Equations • Subtract the new equation from 1st eq • 1.75 Mcal/kg A + 2.45 Mcal/kg B = 10.25 Mcal • 1.75A + 0.865B = 5.45 kg new equation • 0 + 1.59B = 4.8 • B = 4.8/1.59 • B = 3.02 kg grain

  14. Simultaneous Equations • Substitute 3.0 for B in either original eq • 0.561 kg = 0.18 A + 0.089 B • 0.561 kg = 0.18 A + 0.089 (3.0) • 0.561 kg = 0.18 A + 0.27 • 0.561 – 0.27 = 0.18 A • 0.29 = 0.18 A • A = 0.29/0.18 = 1.61 kg alfalfa hay

  15. Simultaneous Equations • Our diet consists of: • 5 kg grass hay • 0.5 kg vegetable oil • 3.02 kg grain • 1.61 kg alfalfa hay

  16. Simultaneous Equations • Check our work: • 5 kg grass hay x 1.95 Mcal/kg = 9.75 Mcal • 0.5 kg vegetable oil x 9.2 Mcal/kg = 4.6 Mcal • 3.02 kg grain x 2.45 Mcal/kg = 7.4 Mcal • 1.61 kg alfalfa hay x 1.75 Mcal/kg = 2.82 Mcal • Do the same for CP 24.6 Mcal

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