The Matheatics 11 Copetency Test Introduction to Unit Conversion: the SI In this the next docuent in this series is presented illustrated an effective reliable approach to carryin out unit conversions startin with a value expressed in one set of units of easureent, calculatin the equivalent quantity in another equivalent set of units of easureent. In this particular docuent, we ll develop the stratey of the ethod by illustratin conversion between units of easureent within the SI. In the next docuent, we ll illustrate how the ethod works for conversions between copatible units in the SI outside the SI. The stratey we will describe here is quite foral it requires you to set your work up in a very specific anner. If you follow the procedure without any shortcuts, the only possible error you can ake is in your arithetic. As we pointed out in an earlier docuent, errors in unit conversion can be very costly, so there is no justification for short-cuts to this ethod, or for usin other less foral less reliable ethods (such as uess--hope-for-the-best, the ost frequently used alternative). Unit Conversion Factors By unit conversion factor, we siply ean any true stateent of the equivalence of two quantities, expressed with different units. Within the SI, these conversion factors coe fro the definition of the base units, the prefixes used to indicate ultiples or fractions of base units, definitions of the various suppleental units. So, for exaple, each of the followin are conversion factors iplied by inforation we ve iven you about the SI: so on. 1 = 100 c 1 in = 60 s 1 k = 1000 1 MW = 1 000 000 W It doesn t really atter how these are stated. So, for instance 1 = 100 c 1 c = 0.01 say exactly the sae thin, are equally useful in principle. One of the biest advantaes of the ethod we are about to describe is that if you know that 1 = 100 c you never have to rearrane this to read 1 c = 0.01 thus one ajor potential source of error in these calculations is eliinated. We stronly recoend that you always use conversion factors in the etric syste which involve whole nubers whenever possible. Tabulations of coonly required unit conversion factors are readily available for units outside the SI. We include a short table in the next docuent in this series. Various publications (for exaple, the CRC Hbook of Cheistry Physics) contain very extensive tables of unit David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 1 of 9
conversion factors. Also, there are a nuber of internet sites listin hundreds of unit conversion factors to find the, use a search enine keywords such as unit conversion, conversion factors, siilar. The Method The ethod consists of, if you like, four siple but essential steps. We will illustrate it with a fairly siple exaple. Exaple 1: Convert the lenth 4.25 k to its equivalent in units of centietres. This is kind of a silly request, it would see, since you iht wonder why we d want to know what such a lare lenth as 4.25 k is in units of centietres. There s probably no ood answer to that question but what you can easily deonstrate is that few people to who this question is posed would be able to state an answer with confidence, so it is a ood exaple to use to illustrate the basic unit conversion ethod. step 1: List a sequence of unit transitions for which you have unit conversion factors, that will et each initial unit in the proble to its correspondin final unit. (This step is always absolutely essential. To skip it is an error because it eans you are willin to accept the risk of a serious error that you could take steps to avoid.) When workin within the SI, use the definitions of the base units prefixes. The sequence of conversions will alost always be siplest to ipleent if it contains the base units. When workin with units outside the SI, you will need to consult an available adequate table of unit conversion factors to develop the stratey which is the oal of this step. At this stae, no nubers are involved (thouh you will have to fill in the nubers eventually). This stae is for developin a plan or stratey only. So, in this exaple, we need a sequence of units that will o fro k to c. Since we know that 1 k = 1000 1 = 100 c an appropriate path is k c Note that since the initial final units here are both in the SI, it has been easy to find a path fro initial to final units that uses the base unit as an interediate. Step 2: Write out a product of the initial quantity to be converted, a factor in the for of a fraction for each arrow that appears in all sequences of conversions developed in step 1. For the oent, each fraction contains just unit sybols in its nuerator denoinator. These unit sybols are placed, top or botto, to cancel the unit at the tail of the arrow, leavin the unit at the head of the arrow. (DO NOT WRITE ANY NUMBERS in these fractions at this step, but leave roo to write in nubers in the next step.) For this exaple, our sequence fro Step 1 has two arrows, so the forula we develop here will have two such fractions: David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 2 of 9
4.25 k = 4.25 k k c Notice: the first fraction here ust involve k (since the first arrow in the sequence fro step 1 involves the conversion fro k to ). The sybol k is placed in the denoinator so that it will cancel the k in the oriinal value. This eans that the sybol ust be placed in the nuerator. the second fraction ust involve c. The sybol is placed in the denoinator since it ust cancel the unit in the nuerator of the previous fraction. This eans that the sybol c ust o in the nuerator. Aain, do not write any nubers in at this stae. Here, we are siply settin up a teplate for the required calculation. By restrictin our attention to the for of the teplate only, we avoid potential confusion error due to tryin to keep too any thins in our inds at once. Notice that if we just look at the unit sybols theselves, cancellation here will occur to produce a result with the desired final units: 4.25 k = 4.25 k k c = c Step : Fill in nubers in the nuerator denoinator of each fraction in the expression fro step 2 so that the nuerator denoinator represent the sae physical quantity. Use your source/list of unit conversion factors to do this. In this exaple, we et 1000 100 c 4.25 k = 4.25 k 1k 1 since we know that 1 k = 1000 1 = 100 c Notice that you use the values in the unit conversion factors exactly as they are iven. It is not ever necessary to rearrane soethin like into 1 = 100 c 1 c = 0.01 first. Notice as well that we didn t have to think too deeply about where each nuber oes, top or botto. Since the first fraction already had the for David W. Sabo (200) Introduction to Unit Conversion: the SI Pae of 9
k since the top botto had to end up bein identical physical quantities (in this case identical lenths), since we know that 1 k = 1000 we know iediately that the 1 oes with the k in the denoinator, the 1000 oes with the in the nuerator. There is no puzzleent whatsoever by this stae of the proble over whether you should ultiply by the 1000 or divide by the 1000 since the teplate developed in step 2 indicates precisely where the factor of 1000 should o in this fraction. This eliinates one ore ajor source of error in unit conversion calculations istakenly ultiplyin when you should divide or dividin when you should ultiply, a coon error when people attept to do unit conversion calculations usin less systeatic ethods. Step 4: Do the nuerical arithetic to et the final answer verify once aain that the units cancel or siplify to ive the desired final units. So, for this exaple, we have 4.25 k = 4.25 k 1000 1 k 100 c 1 = ( 4.25) ( 1000) ( 100) c = 42 500 c as the final answer. This first exaple appears to be very lenthy because we described the ethod discussed issues strateies in considerable detail as we worked throuh the actual exaple. Now we ll repeat this exaple without all of the discussion. Exaple 1 (Repeat): Convert the lenth 4.25 k to its equivalent in units of centietres. The required conversion here is fro kiloetres to centietres. We know that 1 k = 1000 1 = 100 c. So (step 1) a plan to accoplish the conversion fro kiloetres to centietres is k c Then (step 2) the required teplate is David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 4 of 9
4.25 k 4.25 k = ( ) k c This teplate indicates that the plan will work, so insert the required nubers fro the conversion factors to et (step ) 4.25 k 4.25 k = ( ) = 1000 1 k ( 4.25)( 1000) ( 100) c 100 c 1 = 42 500 c as the final answer. You ve probably noticed that what akes this stratey work is that we are siply ultiplyin the oriinal value by fractions with equivalent nuerators denoinators (hence they are equal to 1 so the ultiplication doesn t chane the real value of the oriinal quantity). However, these fractions ultiplied onto the oriinal value are set up to cause undesired units to cancel to leave the desired units in their place. Exaple 2: Convert the speed of 550 c/s to its equivalent in k/h. This is a ore difficult proble than the first exaple. Obviously we need to convert c to k, we need to convert s to h. Now since we know that 100 c = 1 1000 = 1 k, as well, we know that 1 in = 60 s 1 h = 60 in, then, strateies we can ipleent here are c k s in h David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 5 of 9
This copletes step 1. Since our strateies include four arrows altoether, step 2 will now require the forulation of four fractions, once for each arrow in the strateies: c c k s in 550 = 550 s s c in h Make sure you underst exactly why this expression is the correct one to write down for step 2. Now, insert the nubers in these fractions appropriately c c 1 1k 60s 60in 550 = 550 s s 100 c 1000 1in 1h = 550 c s 1 100 c 1k 60 s 60 in 1000 1in 1h = ( 550)( 60)( 60) ( 100)( 1000) k h = 19.8 k h Thus, the speed of 550 c/s is equivalent to 19.8 k/h. Exaple : The concentration of salt in a solution is 0.17 /c. Convert this concentration to units of k/. You ay use the facts that 1 = 1000 litres 1 litre = 1000 c. In this conversion, we will be convertin the ass units of to k, the volue units of c to. The ost obvious strateies, iven the conversion factors stated in the proble for volues the fact that we know that are 1 k = 1000 k c litre. Thus, step 2 ives the teplate k c litre 0.17 = 0.17 c c litre which becoes (step ) David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 6 of 9
0.17 0.17 c = c 1k 1000 c 1000 1 litre 1000 litre 1 = ( ) ( ) ( 1000) ( 1000 ) 0.17 1000 k = 17 k as the final answer. Exaple 4: Repeat the conversion in Exaple above, but use a conversion factor relatin centietres to etres directly. We are asked to do 0.17? c k as in Exaple. Now, however, instead of the stratey c litre we are asked to use siply c. As before, the stratey for the ass unit conversion will be siply k Now, in step 2, we ay start out writin k c 0.17 = 0.17 c c But, if you check, you ll see that this last fraction replaces only one of the three centietre units by a unit of etres. In fact, we ust repeat the last factor three ties to coplete the required unit conversion: 0.17 0.17 c = c k c c c = k We could copact this notation a bit by usin a power or exponent notation: David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 7 of 9
0.17 0.17 c = c k c keepin in ind that because of the exponent, the c in the nuerator of the last fraction really aounts to c repeated three ties. Now, we can fill in the blanks with the appropriate nubers, do the arithetic to et the final answer: 1k 100 c 0.17 = 0.17 c c 1000 1 as obtained before. = 0.17 c ( 1k)( 100 ) ( 1000 )( ) c 1 100 k = ( 0.17) = 17 1000 k Reeber that when a bracketed expression is raised to a power, every factor in the expression inside those brackets is raised to the sae power. In detail, this eans that 100 c 100 c c 1 = = 1 100 Exaple 5: A rectanular field is 185 lon 17 wide. Copute its area state your answer in units of hectares. First, to et the area of a rectanular reion, we need to ultiply the lenth by the width. Here, this ives A = LW = (185 )(17 ) = 2545 2. Now we need to convert this value to units of hectares. Lookin back in the precedin docuent in this series, we find that 1 hectare = 10 000 2. Thus, 2545 = 2545 2 2 i 1hectare 10 000 2 David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 8 of 9
= 2.545 hectares The area of this rectanular field is 2.545 hectares. David W. Sabo (200) Introduction to Unit Conversion: the SI Pae 9 of 9