[Guardsman]

Hey guys, I want to go to school for engineering, Electrical or Electronics (both, if they have a program like that).

Anyhow, I have a big problem that currently prevents me from finishing even my Associates (Community College of the Air Force).....I can't do text-book algebra.

I can do applied algebra, I've been doing it since I was a kid. But, put a text book in front of me, with x=axb+c2/d, and I might as well be reading Latin.

One test study guide that I looked at had a question asking what the square root of i is. I can't even begin to understand how I get an answer to that. My line of thinking is that an imaginary number doesn't exist, therefore, it can't have a square root. Nothing = nothing.

I worked at Stanford Linear Accelerator Center until recently, and I looked over some of the papers and machine technical stuff, in how the accelerator works, beam physics, etc.

When I look at that kind of stuff, it makes sense to me....I may not know the exact answer, but if I had all of the necessary information, I could figure out the answer.

I also picked up a book, "Engineering Formulas", by two German guys, and looking through there, the formulas make sense to me.

Has anybody had this kind of problem before, or know somebody that did, or just have any suggestions that might help me out?

I love doing math, but it frustrates the crap out of me that I can't get past this. I tried taking algebra twice in high school, flunked both times, tried it again in 1998, and got a D on that.

I don't want to just pass the classes, I actually want to learn the stuff. Any help would be appreciated. Thanks.

John

[sgarnett]

One thing that I find very useful when working through a complicated "application" problem is to be very rigorous with the units. It can help keep you on track and validate the answer. For example, if you are expecting an answer in pounds (or whatever) and you end up with inches, you made a mistake along the way. Looking back through your steps, everything after the last instance of "pounds" canceled out is wrong. The actual mistake may be even earlier, but that will help narrow it down. I also helps keep things making sense along the way.

Let's say you want to know how far (in feet) the of the car will rise while braking if 400 pounds is transferred to the front wheels. Let's assume the effective spring rate at the wheels is 100 pounds/inch. There are two springs acting in parallel, one per side.

RISE(feet) = (400(pounds) / (2 * 100(pounds/inch))) * (1(feet))/12(inches))
RISE(feet) = (400/200) * (pounds / (pounds/inch)) * (1/12) * (foot/inches)
RISE(feet) = 2 * (1/12) * (inch) * (foot/inches)
RISE(feet) = (2/12) (feet)
RISE(feet) = 0.167 (feet)

First of all, notice that the units match on both sides of the final equation. That's a good sign that the answer might be right. Second, it's early and I haven'r had much coffee. I did NOT just recite that first equation from memory - I used the units. I knew if I was starting with a force in pounds and a rate (also known as a derivative) in force per distance and wanted to end up with distance, then I had to divide so that the units of force would cancel. Similiarly, looking at the units tells me whether I need to multiply the result in inches by feet/inches or inches/feet. If I want inches to cancel and leave feet, it better be (inches) / (inches/feet). Without the units, it's easy to multiply when you should divide, etc.

OK, that's a pretty simple example. You probably coulds arrive at the same answer with just a calculator and common sense. However, note two things:

1) The units are treated as actual terms in the equations. They follow the same rules.
2) The units look an awful lot like the symbols in textbook algebra, don't they?

In fact, the following is perfectly valid:

feet = (pounds / (pounds/inch)) * (feet/inches)
feet = (pound/pound) * inch * (feet/inches)
feet = (inch/inch) * feet
feet = feet

So, the practical application isn't really all that different from the "textbook" stuff.