53b35c1e841b6ba987ba58493259e1a4.ppt

- Количество слайдов: 25

Estimation • Significant Digits (Significant Figures) • Accuracy and Precision • Approximations

Significant Digits • Any digit used in writing a number representing a quantity or magnitude other than those zeros that are used to locate a decimal point or that have a non-zero digit to their left • It is standard practice to count the first doubtful digit as significant

Rules for Significant Digits Multiplication and Division • The product or quotient should contain the same number of significant digits as those contained in the number with the fewest significant digits. • e. g. , (2. 43)(17. 675)=42. 95025 should be expressed as 43. 0 (3 significant digits, same as 2. 43, not 7 -the actual product)

More Examples • Using an exact conversion factor (2. 479 hr)(60 min/hr)=148. 74 minutes (5 SF? ) Express the answer as 148. 7 minutes (4 SF, same as in the number 2. 479) • Conversion factor not exact (4. 00 x 102 kg)(2. 2046 lbm/kg)=881. 84 lbm (5 SF? ) Express the answer as 882 lbm (3 SF as in 4. 00 x 102 kg)

One More… • Quotient 589. 62/1. 246=473. 21027 (Should this be 8 SF? ) Express the answer as 473. 2 which is correct to 4 SF, the number of SF in 1. 246)

Addition and Subtraction • Show significant digits only as far to the right as is seen in the least precise number in the calculation (the last number may be an estimate). 1725. 463 189. 2 (least precise) 16. 73 1931. 393 Report as 1931. 4

More on Addition and Subtraction 897. 0 - 0. 0922 896. 9078 <- less precise <- more precise Report as 896. 9

Combined Operations • When adding products or quotients, perform the multiplication/division first, establish the correct number of significant figures, and then add/subtract and round properly. • If results of additions/subtractions are to be multiplied/divided, determine significant figures as operations are performed. If using a calculator, report a reasonable number of significant figures.

Rules for Rounding • Increase the last digit by 1 if the first digit dropped is 5 or greater 827. 48 becomes 827. 5 for 4 SF 827. 48 becomes 827 for 3 SF 23. 650 becomes 23. 7 for 3 SF 0. 0143 becomes 0. 014 for 2 SF

Accuracy and Precision • Accuracy is the measure of the nearness of a given value to the correct or true value. • Precision is the repeatability of a measurement, i. e. , how close successive measurements are to each other. • Accuracy can be expressed as a range of values around the true value, usually shown as a value with a +/- range. 32. 3+0. 2 means that the true value lies between 32. 1 and 32. 5

Accuracy and Precision • The range of a permissible error can also be expressed as a percentage of the value. Consider a thermometer where the accuracy is given as + 1% of full scale. If the full scale reading is 220 o. F then readings should be within + 2. 2 o of the true value, i. e. , 220 x 0. 01=2. 2

Approximations • Precision is a desirable attribute of engineering work • You do not always have time to be precise • You need to be able to estimate (approximate) an answer to a given problem within tight time and cost constraints.

Approximations • A civil engineer is asked to estimate the amount of land required for a landfill. This landfill will need to operate for the coming ten years for a city of 12000 people. • How would you approach this estimation problem?

Approximations • The engineer knows that the national average solid waste production is 2. 75 kg person per day. He then estimates that each person will generate (2. 75 kg/day)(365 days/year) = 1000 kg/year • The engineer’s experience with landfills says that refuse can be compacted to 400600 kg/m 3.

Approximations • This leads to the conclusion that the person landfill volume will be 2 m 3 per year. • One acre filled 1 m deep will hold one year’s refuse of 2000 people. (We get this from 1 acre =4047 m 2). • The area requirement would then be 1 acre filled to a depth of 6 meters.

Approximations • But the engineer knows that bedrock exists at the proposed site at a depth of 6 meters. So the estimated depth needs to be reduced to 4 meters and the area needs to be increased to 1. 5 acres for 1 year, or 15 acres for a 10 year landfill life.

Approximations • To allow for expected population growth the engineer revises the final estimate to 20 acres for a landfill life of 10 years.

Now It’s Your Turn… • Estimate the cost to launch a communications satellite. The satellite should have a life of 12 years. • The satellite has 24 transponders plus 6 spares that weigh 12 pounds each.

Communications Satellite Each transponder requires: • 20 lbs. of avionics • 40 lbs. of batteries and solar cells The satellite uses 80 pounds of stationkeeping fuel per year The satellite carries an apogee kick motor that weighs 3000 lbs.

Launch Vehicle • Cost to launch on a Delta rocket is $8000/lb. per lb. up to 6000 lb. and $10000/lb. for each pound over 6000 lbs. • Cost to launch on an Atlas-Centaur rocket is $9000 per lb. • Which is the more economical launch vehicle for this spacecraft?

Solution • 24 transponders plus 6 spares at 12 lbs. each weighs 360 lbs. • 20 lbs. of avionics per transponder (30) weighs 600 lbs. • 40 lbs. of batteries and solar cells per transponder (30) weighs 1200 lbs. • 80 lbs. of station-keeping fuel per year (12) weighs 960 lbs.

Spacecraft Total Weight • • Transponders Avionics Batteries and solar cells Station -keeping fuel Spacecraft weight Apogee kick motor Total weight at launch 360 lbs. 600 lbs. 1200 lbs 960 lbs. 3120 lbs. 3000 lbs. 6120 lbs.

Launch Costs • For Delta: (6000 lbs. )($8000) +(120 lbs. )($10000) = $49. 2 M • For Atlas-Centaur: (6120 lbs. )($9000) = $55. 08 M • Launching on Delta is cheaper by $5. 88 M

Reading A Scale Measure is to read between the scale division marks. That is, to estimate the measurement to the nearest one tenth of the space between scale divisions

Reading A Length What is the length of the beetle‘s body? _ Between 0 and 2 in _ Between 1. 5 and 1. 6 in _ Between 1. 54 and 1. 56 in _ Between 1. 546 and 1. 547 in