Improve Conversions: Struggling with Engineering Math? Get Expert Help Now!

[Tyrion101]

This is just a general cry for help. I've recently begun my actual engineering classes, and we started off with perhaps my weakest subject in mathematics, conversion. I can do fine if it's just meters to feet, or miles per hour to feet per second, but what do you do when you need say an answer in feet, but you have things like kilograms/meters^3, or other extraneous bits and pieces left over?

 

[jedishrfu]

There are some good video tutorials online on "Unit Conversions" and I found this discussion that may help:

http://www.purplemath.com/modules/units2.htm

 

[SteamKing]

This is just a general cry for help. I've recently begun my actual engineering classes, and we started off with perhaps my weakest subject in mathematics, conversion. I can do fine if it's just meters to feet, or miles per hour to feet per second, but what do you do when you need say an answer in feet, but you have things like kilograms/meters^3, or other extraneous bits and pieces left over?

If you want an answer in feet, but you are starting with kg/m³, then you probably got something wrong to start with.

If you don't thoroughly understand the two major unit systems in use, SI and Imperial, your engineering career will be brief and disappointing. Even assuming you will do most of your course work in SI, there are many derived units contained in that system which the user should thoroughly understand.

 

[Mark44]

This is just a general cry for help. I've recently begun my actual engineering classes, and we started off with perhaps my weakest subject in mathematics, conversion. I can do fine if it's just meters to feet, or miles per hour to feet per second, but what do you do when you need say an answer in feet, but you have things like kilograms/meters^3, or other extraneous bits and pieces left over?

If your answer should be in units of feet, but you have kg/m³ left over, then you have done something wrong.

A very important concept in converting between two different measurements is that you can always multiply by 1, in some form, to go from one measurement unit to another. Every conversion formula can be used to yield "1" is some form. For example, 1 m. = 39.37 in. If we divide both sides by "m." we get $1 = 39.37 \frac{in}{m}$. We can also divide the given equation by "in." to get $\frac{1}{39.37} \frac{m}{in} = 1$.

I can use these ideas to convert 6 yards to the equivalent number of centimeters.

$6 yd = 6 yd * 3 \frac{ft}{yd} * 12 \frac{in}{ft} * \frac{1}{39.37} \frac{m}{in} * 100 \frac{cm}{m}$
In the expression on the right, each factor after the first represents "1" in some form. It's always valid to multiply by 1, as it won't change the underlying value of the thing being multiplied. As you go across the expression on the right, notice that the yd units cancel, the ft units cancel, the in units cancel, and the m units cancel, and we're left with cm units on the right. For the numeric part, just carry out the multiplication: 6 * 3 * 12 * 1/39.37 * 100, which is about 548.64, in units of centimeters.

As a quick sanity check, a yard is about a meter, so 6 yards should about 6 meters, or a little less (a meter is longer than a yard). Our answer of about 549 cm is 5.59 m, so this answer passes the smell test.

Edit: It would be helpful, Tyrion, if you posted a specific example of where you're having trouble, instead of a generalized question.

 

[Windadct]

One point I will make - ALWAYS include the units in the mathematical formula as Mark has done... the Units should match or cancel for every step.

We had a habit of converting all answers to Furlongs per Fortnight in physics if the Prof or TA did not designate the necessary units for an answer ( for a velocity or speed of course)...

 

[Mark44]

One point I will make - ALWAYS include the units in the mathematical formula as Mark has done... the Units should match or cancel for every step.

We had a habit of converting all answers to Furlongs per Fortnight in physics if the Prof or TA did not designate the necessary units for an answer ( for a velocity or speed of course)...

For extra credit, go from Avogadro's number ($\frac{\text{atoms}}{\text{mole}}$) to $\frac{\text{electrons}}{\text{pint}}$