
Team Number: 079 School Name: Manzano High School Area of Science: Mathematics Project Title: Is there a Santa Claus? Project Abstract: http://mode.lanl.k12.nm.us/97.98/abstracts/079.html Interim Report: http://mode.lanl.k12.nm.us/97.98/interims/079.html Final Report: http://mode.lanl.k12.nm.us/97.98/finalreports/079/finalreport.html
We are writing you in regards to the NMSCC project. Included with the following is our midterm report which is due on 11398.
In our project, we have Found many factors necessary to determine if Santa Claus Flight around the world on Christmas Eve could actually concur. In the following, we will give you some of theitems we personally have concluded:
total land area of earth= 1.76 * 10^14 meters^2
Santa would only have a very maximum of 24.0 hours to cover 5.14*10^14 square meters, of which only 1/3 is actually land (1.76*10^14)
To do so, he will only have 1.00 hours to travel to every household in a single time zone & proceed to the next time zone to the west.
This would be best accomplished by starting with the first time zone to the west of the international dateline (west of the 180 degree longitude line, and progressively moving west until reaching the 180 longitude line again).
Beginning at the North pole, Santa would travel progressively South by Southwest then North by Northwest when changing timezones, and fluctuate between those directions. Assuming that all children are not naughty, but nice (for once) and that all children are asleep, Santa would have to give a gift to each child *****
number of children in the world= 2.00*10^9 *****
Average area to each child = 2.57*10^5 square meters per child (not necessarily true, though) *****
Area of a timezone=2.14*10^13 square meters *****
Average number of children in a timezone =8.33*10^7 children per timezone *****
So, therefore, Santa mush visit 8.33*10^7 children (Children are defined as persons under 18.0 years old) in one block area of 2.14*10^13 square meters large in one hour, each hour on christmas eve. Assuming that each child is equidistantly distributed from eachother, the entire trip to every child in the world would equal (also assuming that no other child will be born on christmas eve) The Number of Children in the World * THE SQUARE ROOT OF the Average Area To Each Child * THE SQUARE ROOT OF 2 *****
Total distance traveled by Santa=1.43*10^12 meters So distance to be traveled by santa is 1.43*10^12 meters. [EXPLANATION: If each child was equidistantly distributed, they'd (in theory) be in a square with area 2.57*10^5. So, by square rooting the area of that square, you'd find one HYPOTHETICAL side of the square (length). By multiplying that by the SQUARE ROOT of that area by the Square root of 2, you'd find the maximum distance each child would be displaced (see geometry book's ISOSCELES TRIANGLE THEOREM). The minimum distance would be ***** = 1.01*10^12. (but because we are trying to be more precise, we will use the larger distance) *****
So santa must travel 1.51*10^3 kilometers per second. *****(comparison)
So santa would (when on sled) not turn into energy and bend the spacetime continuum and travel into the future due to traveling faster than the speed of light. However, Santa would have to travel faster than his sled in order to get off the sled, go down the chiminey, deliver the gifts, and eat what goodies have been left for him (but that has not been calculated in this research, but is hypothesized to be faster than the speed of light, which would do all of the a fore mentioned statement) *****
Speed also = 2.31*10^4 children per second = 4.32*10^5 seconds per child = .432 microseconds per child = 432 nanoseconds per child *****(comparison)
Santa's speed is much faster than the speed of sound, and thus, he will make a sonic boom each time he reaches a house and leaves a house. Also, in traveling faster than the speed of sound, he can not yell, "On Dancer, Prancer, Rudolf, Cupid, Vixen, Comet, Dasher, Donner, and Blitzen!" because he would have to leave before his voice could catch up with them.
Mass of gifts per child= 4.00 kilograms per child [EXPLANATION: a child wouldn't want a gift that he/she can't durably cary, so in estimation, a child wouldn't want a gift more than 4.00kg (in case if a child wants an abstract gift, Santa would give him/her energy to reach for his/her dream or ideal (which in turn, energy from 4 kg of highly concentrated calories could last a very long time)).] *****
So each of santa's reindeer, Santa, his sled, and the gifts would total 8.00*10^9 kilograms *****
Total force exerted by each reindeer= 6.82*10^19 newtons (out of a combined total of 6.14*10^18 newtons by all 9 reindeer)
In conclusion, Santa would have to travel very fast; faster than humanly possible. His reindeer would have to exert forces far beyond those of mortal reindeer. Each child would be woken up by the sonic booms set off by his sled, in which Santa couldn't give them their presents (because they are awake). This summation does not factor in air resistance, time needed to accelerate and decelerate, aerodynamics of sled for optimal output, room for all the gifts, wind, gravity, obstruction of objects in path, corriolis effect, children that live underground, average altitude of earth, fatigue, earth compression, time to feed reindeer, feeding of reindeer and Santa, time for reindeer to excreete "stuff", or time for reindeer games, BUT, somehow, he does it every year.