FEBE1004A – EAD 1B Lecture Packages – Estimations and Approximations PDF

Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg FEBE1004A – EAD 1B Lecture Packages – Package #2 Estimations and Approximations (3 examples) Objectives: To demonstrate the importance and...

Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg FEBE1004A – EAD 1B Lecture Packages – Package #2 Estimations and Approximations (3 examples) Objectives: To demonstrate the importance and effectiveness of back-of-an-envelope calculations. To quickly guesstimate an answer that can be used as an initial estimate, or as an alarm-bell when double-checking a solution in an exam, design, proposal, etc. To highlight the relevance of significant figures. To demonstrate the meanings of: = equals (exactly), ≈ equals (approximately), ≡ is equivalent to, ~ more or less (about), ± positive and negative Note: these guesstimates are based on experiences / general knowledge / intuition / imagination / common sense / …. i.e. there is a subjective component and therefore there is no unique method. Example 1 – Diameter of the Earth Problem 1.) Estimate the Earth's circumference and therefore diameter. statement: 2.) Estimate how fast the Sun moves through the sky. 3.) Estimate the sunset time in Cape Town if it sets at 17h30 in Johannesburg. Assumptions: a.) From experience: the flying time from Johannesburg to London is ~10 hours. b.) Using imagination of a world globe from school: London is about a quarter of the way around the globe, from Johannesburg. Page 1 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 1 – Diameter of the Earth c.) An airliner flies at ~1 000 km/h (limited by the speed of sound, mach 1). d.) London is directly north of Johannesburg (i.e. in-line with each other). e.) The flying time assumes constant speed from point-to-point. f.) The circumference calculated is equivalent to that at the equator. g.) The Sun passes directly over-head South Africa. Methodology: 1.) Earth's circumference and diameter: The distance from Johannesburg to London: distance = speed x time ≈ 1 000 km/h x 10 h ≈ 10 000 km The angular difference between Johannesburg and London: a quarter of a circle ≡ 90° therefore, circumference (360°) of Earth ≈ 4 x distance Johannesburg to London ≈ 40 000 km Diameter of the Earth: Ø = circumference / π ≈ 12 700 km 2.) Speed of Sun through sky: The Earth completes one revolution in 24 hours: speed of Sun through sky ≈ 40 000 km / 24 h ≈ 1 700 km/h Page 2 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 1 – Diameter of the Earth 3.) Sunset time in Cape Town: The angular difference in position between Cape Town and Johannesburg: distance from CT to JHB ≈ 1 000 km longitudinal difference ≈ 1 000 km / 40 000 km x 360° ≈ 9° The Earth's rotational speed is: ω = 360° / 24 h ≡ 15° per hour (towards the east) therefore, time difference = 9° / 15° per hour ≈ 0.6 hours ≡ 36 minutes therefore, sunset time = 17h30 + 0h36 = 18h06 Actual answers: 1.) Circumference = 40 075 km and diameter = 12 756.3 km 2.) Speed of Sun through sky = 1 669.8 km/h 3.) Sunset time = 17h53 Discussion: a.) The actual flying time to London = 11h15. b.) Latitude difference between London (52° N) and Johannesburg (26° S) = 78°. c.) Cruising speed of a Boeing 747 = 933 km/h. d.) London (0° E) and Johannesburg (28°E) are horizontally separated by about 28° / 360° ≈ 8% of the circumference ≡ 3 200 km. This makes the actual distance = 10 496 km (a triangle of ~10 000 km vertically and ~3 200 km horizontally). e.) The actual flying time includes about an hour of manoeuvring at both ends of the journey and is therefore not point-to-point. f.) The Earth is somewhat elliptical and not round. Page 3 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 1 – Diameter of the Earth Recommend- Using the following data would improve the estimates: ations: The actual angular difference (78°) and not quarter-of-globe estimate (90°). The actual distance (10 496 km) and speed flown (933 km/h). The circumference of the Earth at the latitude of Cape Town (less than the equator). Consideration of the latitude of South Africa; in winter, the Sun passes overhead the Tropic of Cancer, ~50° north of South Africa. Cape Town is further south than Johannesburg so the horizontal component is less than 1 000 km. Conclusions: Despite the somewhat inaccurate assumptions and guesstimates, the answers correlate surprisingly well, within 10% ! Testing of one's assumptions is very important to check that they are indeed valid and realistic. Alternatively, they must be refined. When estimating, generally the first digit and the order of magnitude of the number are only meaningful; decimal points are rarely considered. Page 4 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 2 – Population Density of South Africa Problem 1.) Estimate the number of people in South Africa. statement: 2.) Estimate the area (square kilometres) of South Africa. 3.) Estimate the average population density of the country. Assumptions: a.) From Eskom's “49M” energy-saving campaign, circa the Soccer World Cup in 2010, there were ~49M people. b.) The number of people living in Johannesburg is ~4M. c.) The entire population lives in one of the major cities; no one lives in the rural areas. d.) All major cities (considered to be cities with access to an airport) are the same size as Johannesburg. c.) The length of the country, from corner to corner, is ~2 000 km. Methodology: 1.) Estimating the number of people in SA: Eskom's 2010 campaign was based on the approximate number of people i.e. population (2010) ≈ 49M people Alternatively: There are 12 major “cities” in South Africa: JHB, PTA, DBN, PMB, CPT, BFN, KIM, PE, ELS, NEL, Polokwane, Soweto Total population ≈ 12 cities x 4M people per city ≈ 48M people Page 5 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 2 – Population Density of South Africa 2.) Estimating the area (square kilometres) of the country: Considering the shape of the country, the outline fits more-or-less into a big square with a diagonal length of ~2 000 km; the top-left and bottom-right corners fall outside the borders / coastline. The four smaller blue squares are each divided into 2 triangles i.e. 8 triangles in total. The country therefore fills ~6 triangles The diagonal length ≈ 2 000 km ≡ base x height ≈ 1 400 km x 1 400 km Area of shaded country ≈ 6 / 8 x 1 400 km x 1 400 km ≈ 1.5M km2 3.) Estimated average population density: Population density = number of people / area Therefore: 48M / 1.5M km2 ≈ 32 people / km2 Actual answers: 1.) Population (2017) = 56.72M people (i.e. growing at 1M per year since 2010) 2.) Area of South Africa = 1.22M km2 3.) Population density = 46.5 people / km2 (Aside: Gauteng > 700 people / km 2; Northern Cape < 4 people / km2) Discussion: a.) Initial population numbers used are out-of-date and somewhat vague. b.) “Number of people in Johannesburg”... Where is the boundary of Johannesburg? c.) The distance from corner-to-corner is based on the road distance (N1) from Cape Town to Musina (1 925 km). d.) Looking at a map of South Africa, the square approximation is not bad at all. e.) Land-locked countries like Lesotho and Swaziland have been included in the area (albeit small) but excluded from the population numbers. Page 6 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 2 – Population Density of South Africa Recommend- Using the following data would improve the estimates: ations: Reducing the area of South Africa by those of Lesotho and Swaziland. Increasing the number of smaller squares (and hence triangles) to better approximate the shape / area. A better source for the number of people? Conclusions: Again, the answers correlate relatively well, within about 20%. The estimated answers depend significantly on the accuracy of the assumptions: rubbish in = rubbish out. Common sense needs to be used to judge the quality of the assumptions as well as the answers themselves. Page 7 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 3 – Text to Megabytes (MB) Problem 1.) Estimate how many words would need to be typed, to reach 1MB of data. statement: 2.) Estimate how many hand-written exam pages this would fill. 3.) Estimate how many MB the Pocket Oxford Dictionary would be, if typed out and saved. Assumptions: a.) Each character (or letter of text) is 1 byte. b.) On average, there are ~5 letters per word. c.) With reasonable hand-writing, one line can accommodate ~10 words. d.) Each exam page comprises ~30 lines to write on. e.) Writing on one side of a page only. Methodology: 1.) How many words in 1MB: Since each character is one byte, the total number of letters in 1MB is: 1MB = 1024 x 1024 bytes = 1 048 576 bytes ≡ 1 048 576 letters Based on the average number of letters per word: total number of words = 1 048 576 letters / 5 letters per word ≈ 210 000 words 2.) Number of hand-written exam pages: The number of hand-written words that fit onto one page of an exam pad: words per page = 30 lines x 10 words per line = 300 words / page number of exam pages = total words / words per page ≈ 700 pages Page 8 of 9 Faculty of Engineering and the Built Environment University of the Witwatersrand, Johannesburg Example 3 – Text to Megabytes (MB) 3.) How many megabytes would the dictionary be: To estimate the total number of words in the dictionary: there are about ~1 000 pages, each page comprises two columns of ~8 words per line and ~60 lines. total number of words ≈ 1 000 000 total number of letters =1 000 000 words x 5 letters per word ≈ 5 000 000. Total number of MB = 5 000 000 bytes / 1 048 576 ≈ 4.8 MB Actual answers: 1.) Based on the number of letters in the average English word, words per MB = 233 017 words. 2.) Same. 3.) Between 4 MB and 6 MB depending on the version and whether punctuation and spaces are considered or not. Discussion: a.) The quality of hand-writing has a significant impact here. b.) In practice, spaces and punctuation are also characters (special letters) too. Recommend- Using the following data would improve the estimates: ations: The actual average English word comprises 4.5 letters. Counting the number of words on one typical page of the dictionary and checking the total number of pages, would make a significant improvement in the estimate (and is much quicker than counting everything single word on every single page). Conclusions: Sometimes a sample (in this case, of the hand-writing or one page of dictionary text) would provide a very accurate reference. Note: There is a subtle difference between “Mega” in Megabytes = 1024 x 1024 bytes and in Megahertz = 1000 x 1000 hertz. Page 9 of 9