Giant Enchilada
!!UPDATE: THE GIANT ENCHILADA FIESTA NO LONGER EXISTS!! Every year during the last weekend in September (Friday,
Saturday, and Sunday), along a portion
of Hadley Avenue, from Solano Drive east to Walnut street, in Las Cruces, NM,
the Whole Enchilada Fiesta takes place. Roberto Estrada of Roberto’s Mexican
Restaurant in
How to cook enchiladas.
Special
equipment used in the making of THE Enchilada was designed by Robert V.
Estrada. The equipment includes the press, the carrying plate, the cooking vat
and the serving plate.
Technique
First, we place the carrying
tray on top of the press; then we put 250 lbs. of the masa
dough on top of it. The dough is then pressed to make the tortilla.
From here, the tortilla is carried by about 14 men to the cooking vat that
contains vegetable oil, which has been heated to 550 degrees (this is a
difficult procedure).
The tortilla is then cooked and when done, it is carried by the same men, and
laid on the serving plate (this is another one of our difficult procedures).
Robert V. Estrada will then ladle on some of the chile
sauce, then he will spread some of the cheese and then
some of the chopped onions.
This marks the completion of the first layer of THE Enchilada. The second and
third tortillas are done in the same manner. Then THE Enchilada is ready to be
served.
It takes approximately two and a half hours, from start to finish, in the
making of the World’s LARGEST Enchilada!
Questions.
* What is the total area of a round enchilada 11 feet in diameter? The area A of a circle is π*r^2, where r is the radius. If the diameter is 11 feet, the radius is 5.5 feet. So A ≈3.14*5.5^2 ≈ 95 square feet.
* What is the circumference of the enchilada? C/d = π, so C = π*d ≈ 3.14*11 feet = 34.5 feet. Let's get some string and lay out a circle on the floor!
* How much will each square foot of the enchilada weigh? Add the weight of all ingredients (except the oil) and divide by 95 sq. ft. (One gallon of red chiles weighs approximately 8 lb.) Here are the ingredients and their amounts:
3 corn tortillas @ 250 pounds.
75 gallons of red chile sauce @ 8 lb/gallon,
50 lb chopped onions,
175 pounds cheese
So we compute
(3*250 + 8*75 + 50 + 175) lb/95 sq. ft. = 1575/95 = 16.5 pounds/sq. ft.
How big is a square foot? (We gave students square inch graph paper and had them cut out and tape together one square foot. It is bigger than they thought!)
* How thick is the enchilada going to be?
One cubic inch of food weights approximately .55 ounce (the density of water).
Thus 16.5 pounds contains 16.5 lb*16oz/lb /.55oz/cu. in. = 480 cubic inches. There are 12*12*12 cubic inches in a cubic foot, and 12*12 square inches in a square foot, so the height is 480 cubic inches/(12^2 square inches) = 3 1/3 inches.
* How many people could it serve?
The total weight is 1575 pounds. Counting 1/4 pound per person, it can serve 6300 people.
* What amount of ingredients would you need to make an enchilada that is 12 feet across?
The surface area of the enchilada goes up by the SQUARE of the diameter. So multiply the amounts of ingredients by (12/11)^2 ≈ 1.2. (Its height is still the same, 3 1/3 inches.)
* Make a model of this giant (11 feet in diameter) enchilada from modeling clay or Play dough in the scale 1 : 24. This means that 1 inch in your model corresponds to 24 inches (= 2 feet) of the real object. So your model should be 5.5 inches across and (3 1/3)/24 ≈ 1/8 inch thick.
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