ENERGY REQUIREMENTS

OF RUMINANTS

• Total digestible nutrients (TDN)– Traditional system to express digestible energy

concentration of feedstuffs– Basis of TDN are physiological fuel values

– TDN, %DM = %DP + %DCF + %DNFE + (2.25 x %DEE)

Nutrient Heat of combustion,

kcal/gm

Heat of combustion of metabolic

products, kcal/gm

Nutrient absorption, %

Physiological fuel value,

kcal/gm

Carbohydrates 4.1 - 98 4.0

Fats 9.45 - 95 9.0

Protein 5.65 1.30 92 4.0

FEED ENERGY SYSTEMS

• Equivalence in energy units– 1 lb TDN = 2000 kcal Digestible Energy– 1 kg TDN = 4400 kcal Digestible Energy

• Limitations of TDN– Limitations with digestion trials

• Errors in chemical analyses• Errors in digestibility trials

– Low feed intake increases digestibility– DMI at 3x maintenance reduces TDN by 8%

– Underestimates or does not include all energy losses in metabolism

• Underestimates energy loss in urine (5%)• Does not include methane gas

– End product of rumen fermentation– 3 – 10% of feed energy

• Does not include:– Work of digestion– Heat of fermentation– Heat of nutrient metabolism

– Overestimates the usable energy value of feeds• Particularly of forages

CALORIC SYSTEM• Energy units

– Calorie (cal)• Amount of heat required to increase the temperature of 1

gm of water from 14.5 to 15.5oC

– Kilocalorie (kcal) = 1000 cal– Megacalorie (Mcal) = 1000 kcal = 1,000,000 cal

• Caloric system subtracts digestion and metabolic losses from the total energy of a feedstuff

CALORIC SYSTEM

Gross Energy

Fecal Losses Digestible Energy

Urine Losses Gaseous Losses

Heat Increment Losses

Metabolizable Energy

Net Energy

Work of Digestion

Heat of Fermentation

Heat of Nutrient Metabolism

Maintenance Retained Energy

Growth

Stored EnergyLactation

COMPARISON OF ENERGY FRACTIONS IN DIFFERENT FEEDSTUFFS

Corn grain

kcal/g

Alfalfa Hay

(midbloom)

kcal/g

Oat Straw

kcal/g

Gross Energy 4.5 4.6 4.7

Digest. Energy 3.92 2.56 2.21

Metab. Energy 3.25 2.10 1.81

NEm 2.24 1.28 0.97

NEg 1.55 0.68 0.42

CALCULATION OF ENERGY VALUES IN BEEF NRC REQUIREMENT PUBLICATION

• DE = .04409 x TDN (%)• ME = DE x 0.82• NEm = 1.37ME-0.138ME2+0.0105ME3-1.12• NEg = 1.42ME-0.174ME2+0.0122ME3-1.65

where units for DE, ME, and NE are Mcal/kg

CALCULATION OF TDN CONCENTRATIONS IN DAIRY NRC REQUIREMENT PUBLICATION

• Inputs• tdNFC = .98(100-[(NDF-NDICP)+CP+EE+ash])xPAF

where PAF = .95 for cracked corn 1.00 for ground corn

1.04 for HM corn.94 for normal corn silage.87 for mature corn silage

• tdCPf = CP x e(-1.2xADICP/CP)

• tdCPc = [1 – (0.4 x (ADICP/CP))] x CP• tdFA = FA• tdNDF = 0.75 x [(NDF-NDICP)-L) x [1 – (L/(NDF-NDICP)).667]• TDN• TDN1x= tdNFC + tdCP + (2.25 x tdFA) + tdNDF – 7• Other specific equations for animal protein supplements and fat supplements

CALCULATION OF ENERGY CONCENTRATIONS IN DAIRY NRC REQUIREMENT PUBLICATION

• DE1x (Mcal/kg) = (tdNFC/100) x 4.2 + (tdNDF/100) x 4.2 + (tdCP/100) x 5.6 + (FA/100) x 9.4 – 0.3

• Intake discount = [(TDN1x – [(0.18 x TDN1x) – 10.3] x Intake)] / TDN1x

where intake is a multiple of maintenance

• MEp (Mcal/kg) = [1.01 x DEp – 0.45] + 0.0046 x (EE – 3)

• NElp

– For feeds with < 3% EE

NElp (Mcal/kg) = [0.703 x MEp] – 0.19

– For feeds with > 3% EE

NElp (Mcal/kg) = [0.703 x MEp] – 0.19 + ([(0.097x MEp + 0.19)/97] x [EE – 3])

DISCOUNT FACTORS FOR TDN FOR RATIONS WITH DIFFERENT TDN1X AT INCREASING

LEVELS OF DM INTAKE

Efficiency of NEEfficiency of NE

Maintenance (60 – 70%)

Lactation (64%)

Growth (25 – 45%)

En

erg

y b

ala

nce

0

-

+

73.5 kcal NE/kg.75

Energy intake

Implications of differences in efficiency of energy use for different functions

• When calculating energy needs– Mature dairy cattle can use one value to express the needs

for maintenance and lactation (NEl)

– Growing cattle must use separate values to express the needs for maintenance (NEm) and gain (NEg)

Energy requirements

• Maintenance– % of total energy requirement

• 25 – 70% in dairy cattle• 70% in beef cattle

– Components• Basal metabolic rate• Activity• Body temperature regulation

• Pregnancy• Growth• Lactation

CALCULATION OF THE MAINTENANCE REQUIREMENTS FOR NET ENERGY

FOR BEEF AND DAIRY CATTLE• Beef cattle

– NEm = 0.077EBW.75

• Dairy cattle– NEl for maintenance = 0.080BW.75

MAINTENANCE MODIFIERS(All except lactation apply across sexes)

• Breed

– Implications• Maintenance requirements of breeds with high milk potential

are 20% higher than those with low milk potential– Maintenance requirements of Bos indicus breeds are 10% lower than

Bos taurus• Maintenance represents 70% of total annual ME requirement

of beef cows– Match cow breeds to feed resources

Maintenance

Breed Kcal ME/BW.75 Mcal/d % of total annual ME

Angus x Hereford

130 14 73

Charolais x 129 15 73

Jersey x 145 14.2 71

Simmental x 160 17.9 75

RELATIONSHIP OF BIOLOGICAL EFFICIENCY AND FEED AVAILABILITY

Maximum DMI at Max DMI, kg/yr___Breed__ efficiency efficiency 3500 7500 gm calf kg/yr gm calf weaned/kg DMI/ weaned/kg DMI/ cow exposed cow exposedRed Poll 47.1 3790 47 24Angus 41.3 4111 39 17Hereford 35.1 4281 30 13Pinzgauer 46.9 5473 38 44Gelvieh 44.5 5475 29 36Braunvieh 39.4 7031 33 42Limousin 39.4 7498 33 42Simmental 41.5 8609 26 42• Effects of feed availability on biological efficiency

– Rebreeding rates– Weaning weights

• Reasons for difference in energy requirements between breeds– Difference in energy expenditure of visceral organs

– Difference in protein and fat turnover• Efficiency of protein accretion = 40%• Efficiency of fat accretion = 60 to 80%

Organ Linear contrast (High milk prod vs low milk prod),

g/kg.75

Heart 1.92

Lung 5.52

Kidney 1.86

Liver 5.83 Blood flow = 25% of cardiac outputO2 consumption = 15% of total

GI tract - Blood flow = 20% of cardiac outputO2 consumption = 11% of total

• Sex– Increase NEm requirement by 15% for bulls

• Lactation – Maintenance requirement of lactating cows is 20%

higher than dry cows– Implications

• Early weaning of beef cows reduces maintenance energy requirement

– Reduces feed use– Stimulates reproduction

Postpartum energy fed

Weaning systems

% cows cycling 60 d post-partum

Low (70% NRC)

Early (38 days) 62.5

Normal (7 mo) 26.7

Medium Early 88.9

Normal 13.3

• Body condition effects– Reflects previous nutrition– NEm = 0.077BW.75 x (.8 + ((CS-1) x .05)

– Implications• Can have compensatory gain in growing cattle or reduce

feed requirements of beef cows by restricting nutrition

• Activity allowance (Beef)– Variation

• 10-20% increase in NEm reqt. for good pasture• 50% increase in NEm reqt on poor hilly pasture.

– Nemact = [(.006 x DMI x (.9 – TDN)) + (.05T/(GF + 3))] x w/4.184

Where DMI is in kg/dTDN is a decimalT is terrain (1=flat, 1.5=undulating, 2=hilly)GF is green forage available in metric ton/ha

• Activity allowance (Dairy)– Walking

• Adjustment = .00045 Mcal Nei/kg BW/horizontal km

– Eating• Adjustment = .0012 Mcal Nei/kg BW • Assumes 60% of diet is pasture

– Walking• Adjustment = .006 Mcal Nei/kg BW• Assumes a hilly pasture is one in which cattle move 200

m of vertical distance/day– Example

% increase in maintenance

Flat, close to parlor, good pasture

Hilly, far from parlor, good pasture

Horizontal movement 2.8 11.4

Eating 7.6 7.6

Terrain - 37.9

Total 10.4 56.9

TEMPERATURE EFFECTS

• Previous temperature– Adjustment

• NEm = (.0007 x (20-Tempprevious) + 0.077) Mcal/BW.75

Body temperature regulation

High

Heat Production

Low

Low Temperature High

Basal Metabolic Rate

‘Well insulated conditions’‘W

et/poor in

sulated

condition

s’

UCT

UCT

UCT

‘normal conditions’

LCT

LCT

LCT

HI

Activity

Normal LCT (Cattle)Fasted 18-20CFed 7C

39 C

Effects in applied nutrition

• Mature dairy cows– Cold stress

• Not considered by NRC• Reasons

– High heat production

– Maintained in confinement

– Heat stress• Increase maintenance NE requirement by 25%

• Beef cattle (and dairy heifers)– Cold stress may have major effects on NEm

requirement– Components

• Surface area = SA = .09BW.67

• External insulation = EI = (7.36 – 0.296 Wind + 2.55 Hair) x Hide x Mud

– Determined by:» Wind» Coat length» Hide thickness» Mud or snow Effects on EI Some mud -20% Wet matted -50% Snow covered -80%

• Internal insulation = II = 5.25 + (.75 x CS)– Adult cattle

Total insulation = TI = EI + II• Diet heat increment = HI = (MEI-NEI)/SA

– Calculations

Lower Critical Temp = LCT = 39 – (TI X HI x 0.85)

NEcold stress = SA (LCT-Current temp)/TI x Diet NEm

Diet ME

Add to NEm requirement for total Nem requirement

Example: 600 kg cow (BCS = 5) with a dry coat at -5C temp.

Cow BW NEm reqt DM intakeBCS Wind Coat length Hide thickness Coat cover Diet ME Diet NE Current tempkg Mcal/day kg km/h cm Factor Factor Mcal/kg Mcal/kg C

600 9.334782 15 5 7.5 2.54 Average (1) Dry (1) 2.68 1.76 -5

Surface area = .09 BW.67 6.540393032

EI= 11.617II=5.25+(.75XCS) 9Total insulation= 20.617

ME intake = DMI X ME conc 40.2NE intake = DMI X NE conc 26.4Heat increment = (MEI-NEI)/SA 2.109964941

Lower critical temp =(39-(TI x HI x .85) 2.024024886NEcold stress = SA x ((LCT-Current temp)/TI) * Diet NE/Diet ME 1.463330052

Example: 600 kg cow (BCS = 5) with a snow-covered coat at -5C temp.

Cow BW NEm reqt DM intakeBCS Wind Coat length Hide thickness Coat cover Diet ME Diet NE Current tempkg Mcal/day kg km/h cm Factor Factor Mcal/kg Mcal/kg C

600 9.334782 15 5 7.5 2.54 Average (1) Snow cov (.2) 2.68 1.76 -5

Surface area = .09 BW.67 6.540393032

EI= 2.3234II=5.25+(.75XCS) 9Total insulation= 11.3234

ME intake = DMI X ME conc 40.2NE intake = DMI X NE conc 26.4Heat increment = (MEI-NEI)/SA 2.109964941

Lower critical temp =(39-(TI x HI x .85) 18.69181954NEcold stress = SA x ((LCT-Current temp)/TI) * Diet NE/Diet ME 8.986762984

• Heat stress in beef cattle– Shallow panting = Increases NEm reqt by 7%– Open mouth panting = Increase NEm reqt by 18%

• Effects of excess protein on NEm requirement– Needed for synthesis of urea above requirements– Calculation

• NEm, Mcal/d = [((rumen N balance, gm – recycle N, gm) +

excess N from MP, gm] x .0113) x NEm/MEdiet

– Not included in NRC beef or dairy requirements• Included in BRANDS and CNCPS program

Pregnancy• Very inefficient utilization of energy (14 to 16%)• Increase energy requirement drastically during last

trimester of gestation Energy reqt in last trimester

% of maintenance

Cattle 180

• Calculations:– Beef

• NEm, kcal/d = 0.576 birth wt (0.4504 – 0.000766t)e(.03233-.0000275t)t

– Dairy• NEl, Mcal/d = [(.00318 x t -.0352) x (birth wt/45)]/.218

Bodyweight gain• Less efficient than maintenance• Calculations

– NEg intake, Mcal/day *= DMI, kg x NEg conc., Mcal/kg• *After maintenance requirement is met

– Shrunk BW, kg = SBW = .96 x Full BW– Standard reference BW = SRW (base = medium-frame

steer)• 478 kg for small marbling• 462 kg for slight marbling• 435 kg for trace marbling

– Equivalent shrunk BW, kg =SBW x SRW/Final SBW– EBW = .891 x EqSBW– EBWG = .956 x SBWG– Retained energy, Mcal/day = .0635 x EBW.75 x EBWG1.097

• Equals NEg intake if known in predicting gain– SBWG, kg = 13.91 x RE.9116 x EqSBW-o.6837

• Adjustments for FSBW • Reduce FSBW by 35 kg if no implant used• Increase FSBW by 35 kg in Trenbolone acetate is used

with an estrogen implant• Increase FSBW by 35 kg if extended periods of slow

rates of grain• Reduce FSBW by 35 kg if fed high energy from weaning

to finish

ExamplePredict the rate of gain of a 700 lb (318 kg) Angus steer (BCS = 5) fed 1.5 kg corn silage, 5 kg corn grain, and 1 kg soybean meal (DM basis) that will finish at 1250 lb

(568 kg) at small marbling with no environmental stress.

• Step 1. Calculate NEm and NEg concentrations of diet

– If fed an ionophore, increase NEm conc by 12%

Feed DMI, kg Mcal/kg Mcal/day Mcal/kg Mcal/dayCorn silage 1.5 x 1.69 2.535 x 1.08 1.62Corn 5 x 2.24 11.2 x 1.55 7.75SBM 1 x 2.06 2.06 x 1.4 1.4

7.5 15.795 10.77Nem, Mcal/kg = 15.80/7.5 = 2.106Neg, Mcal/kg = 10.77/7.5= 1.436

Nem Neg

• Step 2. Calculate feed required for maintenance

Nem reqt, Mcal/day = .077 BW.75 x (.8 + ((BCS-1)X.05))

BW BCS Mcal Nem/day318 5 5.798439

Feed for maintenance= Nem reqt/Nem concNem reqt Nem conc Feed for maintenance

5.798439 2.106 2.753295 kg/day

No adjustments for breed or temperature stress needed in this problem

• Step 3. Calculate NEg remaining for gain

Feed for gain = DMI - Feed for maintenanceTotal DMI Feedmaint Feedgain

7.5 2.753295 4.746705 kg/day

Neg intake above maintenance =Feedgain X Neg conc

Feedgain Neg conc Neg intake above maintenance4.746705 1.436 6.816269 Mcal/day

• Step 4. Calculate the Equivalent Shrunk BWSBW = .96 x Full BW

Full BW SBWCurrent 318 305.28Finish 568 545.28

EqSBW = CSBW x SRW/FSBWCSBW SRW FSBW EqSBW

305.28 478 545.28 267.6127

• Step 5. Calculate Shrunk BW gain

SBWG= 13.91 x NEgI..9116x EqSBW-.6837

NEgI EqSBW6.816269 267.6127 1.75187 kg/day

Lactation(Dairy)

• Equal efficiency to maintenance

• NEl reqt for lactation, Mcal/day = kg milk/day x

(.0929 x % milk fat) + (.0547 x % milk protein/.93) + (.0395 x % lactose)– Simply add to NEl needed for maintenance

• Energy from body tissue loss (5-point BCS scale)

Body condition score Mcal NEl/kg BW loss

2 3.83

2.5 4.29

3 4.68

3.5 5.10

4 5.57

ExampleHow much milk with a composition of 3.5% fat, 3.3%

protein, and 5% lactose should a 1450 lb (659 kg) Holstein cow produce if she is consuming a diet

containing 2 kg alfalfa hay, 5 kg alfalfa haylage, 5 kg corn silage, 10 kg corn grain, and 2 kg soybean meal

(DM basis)?

• Step 1. Calculate the NEl intake

DMI Nel conc NEIFeed kg Mcal/kg Mcal/dayAlfalfa hay 2 1.38 2.76Alfalfa haylage 5 1.34 6.7Corn silage 5 1.45 7.25Corn 10 2.01 20.1SBM 2 2.21 4.42Total 41.23 Mcal

• Step 2. Calculate the amount of NEl remaining after meeting the maintenance requirement

Maintenance reqt, Mcal = .08 x BW.75

BW Maintenance reqt659 10.40529 Mcal

Nel remaining for milk production = Total Nel - Nel for maintenanceTotal NEL Maint Nel Nel for milk

41.23 10.40529 30.82471 Mcal

• Step 3. Calculate energy concentration in milk

• Step 4. Calculate milk production

Energy requirment for milk, Mcal/kg = (.0929 x Milk fat) + (.0547 x milk protein/.93) + (.0395 x lactose)Milk fat Milk proteinLactose Nel reqt.

3.5 3.3 5 0.716747 Mcal/kg milk

Milk production = Nel for milk/(mcal/kg)Nel for milkMcal/kg milk

30.82471 0.716747 43.00642 kg

Dairy example 2If previous cow was producing 50 kg/day of milk with the given composition, how much tissue would she

need to mobilize at a BCS of 3.5?• Step 1. Calculate total NEl reqt.

Total Nel requirement, Mcal/d = Maintenance + LactationMaintenance reqt, Mcal = .08 x BW.75

BW Maintenance reqt659 10.40529 Mcal

Lactation reqt, Mcal = kg milk/day x Mcal/kg milkKg milk Mcal/kg milk

50 0.716747 35.83734Total Nel reqt 46.24263 Mcal

• Step 2. Calculate the energy deficit• Step 3. Calculate the amount of tissue needed to fill

deficit

Energy Deficient = NE reqd - NE intakeNE reqd NE intake

46.24263 41.23 5.012625 Mcal

Tissue mobilized= NE deficit / NE conc in tissueNE deficit NE conc, Mcal/kg

5.012625 5.1 0.982868 kg/day

Lactation (Beef)

• Equations– k = 1/T

• T = week of peak lactation

– a = 1/(Peakyld x k x exp)• Peakyld = peak yield, kg/day

– Milk prod, kg/d= Yn = n/(a x expkn)• n = current week

– E = .092 x MF + .049 x SNF - .0509• E = Milk energy, Mcal/kg• MF = Milk fat, %• SNF = Solids not fat, %

– NEm, Mcal/day = Yn x E

FEEDING TO MAINTAIN REPRODUCTION

• Maintaining reproductive performance requires given levels of body fat– No less than 15.8% carcass lipid or 13.5% empty body fat at

parturition• Can be as low as 12.4% empty body fat at parturition if fed at

130% of NRC energy requirement for 60 days post-partum– Empty body fat at breeding should be 15% for optimal

pregnancy rates– Cows should not exceed 20% carcass lipid or 17.8% empty

body fat• Body weight

– Although NRC publications prior to 1996 used body weight, most producers don’t weigh cows

– Body weights of pregnant cows can be confounded with conceptus

USE OF CONDITION SCORING FOR BEEF COWS

• Systems– 9-point visual system (NRC/Oklahoma)– 9-point palpation system (Tennessee)– 5-point visual system (Purdue)

• Limitations – All systems are subjective– Different systems make it difficult to standardize relative to

nutrient requirements• Advantages

– Don’t require weighing of cows– Less confounded by pregnancy than body weights– Related to body weight

• Relationship with BW change– Purdue 1 BCS unit change = 68 kg (5-point system)– NRC 1 BCS unit change = 50 kg (9-point system)

• Relationship varies with age– Mature cows 1 BCS unit change = 34 kg (9-point system)– Primiparous heifers 1 BCS unit change = 68 kg (9 point system)

• Relationship of body condition score to body compositionComponent BCS Change/BCS (5-point) BCS + BW r rCarcass lipid .63 .70Carcass protein .36 .59Empty body lipid .48 5.5-66% units .74Empty body protein .26 .2-1% units .47Hot carcass weight .95Backfat .075-.29 cm .62• Relationship of BCS from different systems to body lipid BCS System9-pt 5-pt NRC, 9-pt. Texas, 9-pt. Purdue, 5-pt Empty body lipid, %1 1 3.77 0 3.12 7.54 4 3 2 11.30 8 8.74 15.07 125 3 18.89 16 14.96 22.61 20 7 4 26.38 24 21.58 30.15 28 9 5 33.91 32 27.2

• Relationship of body condition score to reproduction– Body condition score at calving is the primary factor related to

reestablishment of cyclic activity in beef cows• Cows that calve at BCS > 5 (9-point system) will exhibit estrus

regardless of post-partum nutrition regime

• Feeding extra energy post-partum to cows that calve at BCS < 4 will increase the percentage of cows exhibiting estrus in a finite breeding season

Richards (1986) Days to first estrus Days to conception 1st service conception

Post-partum Calving BCS

nutrition < 4 > 5 < 4 >5 < 4 > 5

High (+.45kg/d) 60 51 91 84 67 59

Mod. ( 0 kg/d) 60 46 91 85 65 67

Low (-.68 kg/d) 56 50 88 82 54 70

L/H (5 kg corn/d 67 49 91 87 75 70

14-d before

and through

breeding)

BODY CONDITION SCORE EFFECTS ON ENERGY RESERVES

• Energy in body condition Body condition score (5-point system) Mcal/kg BW

change

1 2.57

2 3.82

3 5.06

4 6.32

5 7.57– The reason for this difference is that weight change at

condition score 1 is 17% fat, but is 77% fat at condition score 5– Implications

• It takes more energy to increase condition score at a higher condition score than a lower condition score

• Loss of body condition at a high body condition provides more energy than loss of body condition at a low body condition score

• Calculation of energy from body reserves– Body composition from BCS

• Proportion of empty body fat = AF = .037683CS• Proportion of empty body protein = AP = .200886 - .0066762CS• Proportion of empty body water = AW = .766637 - .034506CS• Proportion of empty body ash = AA = .078982 - .00438CS• Empty body weight, kg = EBW = .851SBW• Total ash, kg = TA = AA x EBW

– Calculation of total fat and protein reserves• AA1 = .074602• AF1 = .037683• AP1 = .194208• EBW1, kg = TA/ AA1

• Total fat, kg = TF = AF x EBW• Total protein, kg = TP = AP x EBW• Total fat1, kg = TF1 = EBW1 x AF1

• Total protein1, kg = TP1 = EBW1 x AP1

– Calculation of mobilizable energy• Mobilizable fat = FM = TF - TF1

• Mobilizable protein = PM = TP – TP1 • Energy reserves, Mcal = ER = 9.4FM + 5.7PM

– During mobilization• 1 Mcal ER substitutes for .8 Mcal of NEm

– During repletion• 1 Mcal NEm will provide 1 Mcal ER

Example 1• If a beef cow with a shrunk BW of 485 kg at a BCS 4 has a NEm requirement of 10.46 Mcal/day

is consuming alfalfa hay with a NEm conc of 1.43 Mcal/day at 10.9 kg/d, how long will it take for this cow to increase to a condition score of 5?

• NEm requirement, Mcal/day = 10.46• NEm fed, Mcal/day =1.43 x 10.9 = 15.59• NEm excess or deficient, Mcal/day = fed-reqt = 5.13

• AF at CS4 =.037683 x 4 = 0.1507• AP at CS4 =.200886-.0066762 x 4 = 0.1742• AA at CS4 =.078982-.00438 x 4 = 0.0615• EBW at CS4 =.851x485 = 412.74• Total ash at any BCS =EBW x AA = 25.3675

• AF at CS5 =.037683 x 5 = 0.1884• AP at CS5 =.200886-.0066762 x 5 = 0.1675• AA at CS5 =.078982-.00438 x 5 = 0.0571• EBW at CS5 =25.3675/.0571 = 444.26

• Total fat at CS4, kg =412.74 x .1507 =62.2000• Total protein at CS4, kg =412.74 x .1742 =71.8993• Total fat at CS5, kg =444.26 x .1884 =83.6986• Total protein at CS5, kg =444.26 x .1675 =74.4136

• Metabolizable fat, kg =83.6986-62.2000 =21.4986• Metabolizable protein, kg =74.4136-71.8993 = 2.5143• Energy reserve needed, Mcal =9.4 x 21.4986 + 5.7 x 2.5143 =216.42• Days to increase to CS5 =216.42 x 1/ 5.13 = 42.19

Example 2•If a beef cow with a shrunk BW of 485 kg at a BCS 4 has a NEm requirement of 10.46 Mcal/day is consuming mature bromegrass hay with a NEm conc of 0.94 Mcal/day at 9.7 kg/d, how long will it take for this cow to decrease to a condition score of 3?•NEm requirement, Mcal/day = 10.46•NEm fed, Mcal/day =0.94 x 9.7 = 9.12•NEm excess or deficient, Mcal/day = fed-reqt = -1.34

•AF at CS4 =.037683 x 4 = 0.1507•AP at CS4 =.200886-.0066762 x 4 = 0.1742•AA at CS4 =.078982-.00438 x 4 = 0.0615•EBW at CS4 =.851x485 = 412.74•Total ash at any BCS =EBW x AA = 25.3675

•AF at CS3 =.037683 x 3 = 0.1130•AP at CS3 =.200886-.0066762 x 3 = 0.1809•AA at CS3 =.078982-.00438 x 3 = 0.0658•EBW at CS3 =25.3675/.0658 = 385.52

•Total fat at CS4, kg =412.74 x .1507 =62.2000•Total protein at CS4, kg =412.74 x .1742 =71.8993•Total fat at CS3, kg =385.52 x .1130 =43.5638•Total protein at CS3, kg =385.52 x .1809 =69.7406

•Metabolizable fat, kg =43.5638-62.2000 =-18.6362•Metabolizable protein, kg =69.7406-71.8993 = -2.1587•Energy reserve lost, Mcal =9.4 x 18.6362 + 5.7 x 2.1587 =-187.48•Days to decrease to CS3 =187.48 x .8/ 1.34 = 111.93