# All Formulas

By Vieui
Nov 17, 2013
817 Words

Fiber Content:

FC = Fiber Value / Dietary Fiber

--------------------------------------- X (n= E.P. wt; how many grams where consumed) 100

For Example:

FC = 70 /100 x 200 = 35grams of fiber.

% RENI:

Actual Nutrient Intake

%RENI = -----------------------------------

RENI

For Example:

VitA:

450 / 350 x 100 = 128.6%

Tannhauser’s (Broca) Method:

For example:

An adult sedentary male who stands 5 feet 5 inches.

Ht: = 5 feet 5 inches = 65.0 inches

= 65.0 inches x 2.54 cm

= 165.1 cm

DBW = 165.1 – 100 = 65.1 kg.

= 65.1 – 6.51 (10% of 65.1)

= 58.59 kg or 59 kg

NDAP Method:

DBW for men: 5 feet tall is 112lbs. Add(subtract) 4lbs for every inch above(below) 5 feet. DBW for women: 5 feetl tall is 106 lbs. Add(subtract) 4lbs for every inch above(below) 5 feet.

For Example:

Male: 5 feet and 5 inches

5 feet= 112 lbs

5 in= 20 lbs

= 132 lbs (60kg)

Adopted Method:

Male and Female (DBW)

5 feet = 106 lbs; for every inch above 5 ft, add 5 lbs.

For Example:

5 feet & 5 inches

5 feet= 106 lbs

5 in= 25 lbs

= 131 lbs (59.5kg)

Desirable BMI:

For men: 22

For women: 20.8

BMI = wt(kg)/H(m)2

Note: A BMI range of 19.0 to 24.9 is considered normal.

For Example:

= 61kg / (1.65m)2

= 61kg / 2.7225

= 22.4

Total Energy Allowance (TEA) (Krause Method):

Activity:kcal/kg DBW/day

Bed rest but mobile (hospital patients)27.5

Sedentary (mostly sitting)30.0

Light (tailor, nurse, physician, jeepney driver)35.0

Moderate (carpenter, painter, heavy housework)40.0

Very Active (swimming, lumberman)45.0

For Example:

Total Energy Allowance (TEA) of a Moderate person weighing 59kg.

= 59 x 40 = 2360 kcal or 9875 kj.

Determining CHO, PRO and Fat:

A. Percentage Distribution:

Carbs- 50-70% of TEA

Pro- 10-15% of TEA

Fat- 20-30% of TEA

For Example:

CHO - 65%

PRO - 15%

FAT - 20%

CHO= 2360 x 0.65 =1534 kcal

PRO= 2360 x 0.15 =354 kcal

FAT= 2360 x 0.20 =372 kcal

B. Kcal to Grams

CHO = 4kcal/gram; PRO = 4 kcal/gram; FAT = 9kcal/gram

For Example:

CHO= 1534 / 4 = 383.5 or 385g

PRO= 354 / 4 = 88.5 or 90g

FAT = 372 / 9 = 41.3 or 40g

Diet Rx: Kcal 2350; CHO 385g; PRO 90g; Fat 40g

Deriving Nutrient Values:

1. Carbohydrates

Formula:

% Carbohydrate = 100 – [%Water + %Protein + %Fat + %Ash]

For Example:

Given: Rice, well-milled, boiled (bigas, maputi, sinaing) A020 Water = 67.6g per 100g E.P.

Protein = 2.1g per 100g E.P.

Fat = 0.2g per 100g E.P.

Ash = 0.4g per 100g E.P.

Solution:

Carbohydrate= 100g – (67.6 + 2.1g + 0.2g + 0.4g)

= 100 – 70.3

= 29.7g per 100g

2. Energy

Formula:

Energy (kcal) = [(4xg protein) + (9 x g fat) + (4 x g carbs)] :4,9,4 are Atwater factors

For Example:

Given:Protein = 2.1g per 100g E.P.

Fat = 0.2g per 100g E.P.

Carbohydrate = 29.7g per 100g E.P.

Solution:

Energy (kcal)= [(4 x 2.1g) + (9 x 0.2g) + (4x 29.7g)]

= 8.4 + 1.8 + 118.8

= 129 kcal per 100g E.P.

3. Kcal to kj conversion

1Kcal = 4.184 kj.

For Example:

= 2350kcal x 4.184 kj = 9832.4 or 9850kj.

4. Retinol Equivalent (RE)

Formula: µ

1 µg retinol = 1 RE

1 µg ß-carotene

------------------------ = 1 RE

6

For Example:

Total RE = 9205 (Chicken Liver) + 135 µg per 100g

-----------------------------

6

= 9205 + 23 = 9228 µg per 100g or 9228 RE per 100g.

5. Niacin Equivalent (NE)

Formula:

Mg Tryptophan

Mg niacin = ------------------------------= Niacin + Niacin values from Tryptophan

60= Niacin Equivalent.

6. Niacin Equivalent from Protein

Formula:

Protein / 6 = Additional Niacin.

For Example:

Niacin from soybean seed, dried (C044)= 2.0mg per 100g.

Protein of soybean seed, dried (C044)= 35.8g per 100g.

35.8g / 6 = 5.97 or 6.0mg

= 2.0mg + 6.0mg

= 8.0mg per 100g

Computing Nutrient Content of Foods/Translating Foods of Different Forms and/or Units into Nutrients

7. Convert As Purchased (AP) to Edible Portion (EP) Weight.

Formula: You need FEL manual

E.P. weight(g) = % E.P.

-------------- x A.P. weight (g)

For Example:

Given:

1 pc. Milkfish= 250g A.P. Weight

E.P. Weight of milkfish= 65% (Get it from FCT)

= 250 x .65 = 162.5 or 162g E.P.

8. Convert Cooked A.P./E.P. weight to its raw E.P. weight equivalent. Formulas:

Raw A.P. weight equivalent= A.P/E.P weight of X “Cooked-to-raw” weight CF of the cooked food cooked food

Raw E.P. weight equivalent= A.P. weight X % E.P. / 100

of the cooked food equivalent

For Example:

Given: Cooked weight of scad, round, fried (G105)= 75g

“Cooked-to-raw” weight CF for scad, round, fried= 1.523

E.P. of scad, round = 49%

No FCT data for scad, round, fried

Solution:

Raw A.P. wt equivalent of 75g scad, round, fried= 75 x 1.523

= 114g raw A.P. wt

Raw E.P. wt equivalent of 114g scad, round, fried = 114g x 49/100

= 56g raw E.P. wt

Another Formula:

Raw weight

--------------------- = “Cooked-to-raw” weight CF

Cooked weight

Cooked weight

--------------------- = “Raw-to-Cooked” weight CF

Raw weight

For Example:

“Cooked-to-raw” and “Raw-to-cooked” weight in CF of mungbean

Given:

Wt (g) of raw mungbean= 50g

Wt (g) of cooked mungbean= 120g

Solution:

“Cooked-to-raw” weight CF = 50 / 120 = 0.42

“Raw-to-cooked” weight CF= 120 / 50 = 2.4

Calculate Nutrient Content of Food/Diet

9. Formulas:

FORMULA 10A:

Nutrient per gram E.P. = nutrient per 100g EP (FCT Value) X gram EP Hh measure Hh Measure ------------------------------------------------

100

FORMULA 10B:

Nutrient per 100 EP = X nutrient per gram E.P. Hh measure

---------------------------- -----------------------------------------------------

100g gram E.P. Hh measure

Example: Protein content of 1 pc Banana, lacatan.

Given: 1 pc banana, lacatan = 60g AP

E.P. of banana, lacatan (E011) = 69%

Protein = 1.4g per 100g E.P.

Solution 1 – Using FORMULA 7

E.P. wt of 60g A.P. Banana, lacatan = 69 X 60g

____

100=41g E.P.

Then using FORMULA 10A:

Protein content per 41g E.P. banana, lacatan = 1.4 / 100 x 41g E.P. = 0.6g