# Dimensions in equations

• posted

I'm reading from an electrical engineering book that some voltage v(t) is given by:

v(t) = sin(1000*t)

where t is the time

Isn't this expression, strictly speaking, incorrect? How can we find the sine of a time quantity? Shouldn't we have a constant in there as well with T^-1 dimension to solve this?

For example:

v(t) = sin(1000*k*t)

where k a constant with units 1/s.

• posted

You worry too much! Allow for some jargon. The "1000" used really 1000 radians per second.

Another example, consider that the term "kilocycles" was in common use for decades when it should have been kilocycles per second. Now, of course, the term hertz is used.

Bill

• posted

You are right, strictly speaking. Some details are being omitted, considering only initiated readers. Regards, LL

• posted

Personally, balancing units in equations has proven to be one of the most useful things that I have ever been taught.

-- Sue

• posted

Units are consistent.

• posted

As Salmon egg says sin(1000t) is fine where the 1000 means 1000 radians/second or 159+ Hz. This gives you sin (quantity in radians) which is correct. Dimensionally a radian is T^-1 Note that the expression should read v(t)=Vmax*sin(1000t) where Vmax is in volts (apparently 1V in this case) if you are worried about dimensional analysis. Possibly you missed something earlier in the book.

• posted

Many theoretical physicists do things like setting the speed of light and the mass of an electron equal to one. That way they do not have to write down endless c's, m's, e's etc. When done the put all these symbols back in to get the final formulas with all the formal symbolism.

During one class given by Richard Feynman, he described this technique. One student commented that by leaving out all these dimensions, you could not use dimensional arguments to check the results. Feynman's comment: "Don't make mistakes."

Bill

• posted

"Don Kelly" wrote in news:K%aXi.174477\$1y4.142590 @pd7urf2no:

• posted

It depends on how the symbols are used. If you say the time t is in seconds, then t has a dimension; if you say the time is t seconds, then t is dimensionless (it just says how many seconds you have). Others have already responded by indicating that the 1000 may be shorthand for 1000 radians per second (in which case t had better have a dimension). Of course, you are still left with the fact that w =

1000*t is in radians, so still has a dimension!

R.G. Vickson

• posted

Yeeks. Looks like a terrible book. The formula should look something like this:

v(t) = (v_0)sin(1000*k*t) where k has to have units of 1/s. I would guess in fact that k is a number in kHz, so that 1000*k has units 1/s.

PD

• posted

: > v(t) = sin(1000*t) : >

: > where t is the time : >

: > Isn't this expression, strictly speaking, incorrect? How can we : > find the sine of a time quantity? Shouldn't we have a constant : > in there as well with T^-1 dimension to solve this? : : It depends on how the symbols are used. If you say the time t is in : seconds, then t has a dimension; if you say the time is t seconds, : then t is dimensionless (it just says how many seconds you have). : Others have already responded by indicating that the 1000 may be : shorthand for 1000 radians per second (in which case t had better have : a dimension). Of course, you are still left with the fact that w = : 1000*t is in radians, so still has a dimension! : : R.G. Vickson : : >

: > For example: : >

: > v(t) = sin(1000*k*t) : >

: > where k a constant with units 1/s. :

v(t) = V.sin(omega.t) It may be that V is what he's missing.

• posted

My goof- Thank you.

• posted

Radians are dimensionless, but they are not unitless. You should still carry the units forward in calculations.

dave y.

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