View Full Version : What process do they use to create formulas?


strategicman
06-25-03, 12:30 AM
I was looking at that formula you guys gave me on the variation of time for a constant moving object (I'm not going to write it because I still don't know how to write the greek letters), and I just started wondering about what process they use to find those formulas. Do a bunch of mathemetitions/physicists (I know my spelling sucks) go into a meeting room and start brainstorming, and doing trial and error, or is there some way of doing certain calculations that shows you what the formula would be. Well, I'm just curious, and thanks for your help!

-strategicman

blackholesun
06-25-03, 07:08 AM
Well one way to come up with formulas is to rigorously experiment and try to find equations to match the data. That's the experimental physics branch. Although most experimental physicist get their data from theoretical physics who try to 'predict' the outcome of an experiment by looking at past work or using the fundamentals of a theory previously theorized.

apendrapew
07-01-03, 05:19 PM
This is one thing that I myself have wondered.. Specifically.. I wonder how the hell scientists came up with Vf^2 = Vi^2 + 2ad

That's just nuts. You'd have to be a hardcore geek to figure something like that out.

AD1
07-01-03, 05:39 PM
I wonder how the hell scientists came up with Vf^2 = Vi^2 + 2ad

What do these terms stand for? What does the equation relate to?

Konek
07-01-03, 08:23 PM
That's a velocity equation. Vf is final velocity, Vi is initial velocity. The "a" stands for acceleration and "d" for distance. This only works at constant acceleration. As how they came up with it, it is the result of combining two other equations (which were empirically determined, by the way).

Konek
07-01-03, 08:25 PM
Found you a link: http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot5

AD1
07-01-03, 08:39 PM
Oh yes. I know what the formula is. It's very basic kinematics. I've alway seen it written as

v<sup>2</sup> = u<sup>2</sup> + 2as

Where v is the final velocity, u is the initial and s is the displacement.

which were empirically determined, by the way

I don't think they were, were they? They just come from the definitions of velocity and acceleration.

Konek
07-01-03, 08:47 PM
Originally posted by AD1
...
which were empirically determined, by the way

I don't think they were, were they? They just come from the definitions of velocity and acceleration.

You're right, you don't need experiments, just those two definitions.

Pete
07-01-03, 11:07 PM
If you want a hardcore geek, try Johannes Kepler.

He went through thousands and thousands of records of astronomical observations (at this time on this night, Mars was at this position, etc etc etc), and came up with three 'laws'.

The first notes that the orbit of a planet or comet about the Sun is an ellipse with the Sun at one focus.

The second notes that a line between a particular planet or comet and the Sun sweeps out equal areas in equal time intervals.

The third notes that the square of any planet or comet's sidereal orbital period is proportional to cube of its average distance from Sun.

James R
07-02-03, 12:15 AM
Let's take the example mentioned before:

v<sup>2</sup> = u<sup>2</sup> + 2as.

This is a formula which allows us to calculate the final speed v of an object, provided we know the initial speed u, the distance s the object travelled and the acceleration a of the object. Here's how it is derived.

The first thing to note is that this formula ONLY applies when the object in question is accelerating at a constant rate. That is our starting point.

Acceleration is defined to be the rate of change of velocity:

a = dv/dt ...(1)

So, for an object accelerating at a constant rate we have

v = at + c

where c is a constant which appears when we do the integration of equation (1). At time t=0, v=u, the initial velocity. This means that c=u, so:

v = u + at ...(2)

Now, the velocity is defined to be the rate of change of displacement of the object:

v = ds/dt ...(3)

Using (2), we have

ds/dt = u + at

Integrating, we get

s = ut + (1/2)at<sup>2</sup> ...(4),

where I have set the initial displacement to zero.

Now, we rearrange equation (2) to get:

t = (v-u)/a ...(5)

Replacing t in equation (4) with (5) gives:

s = u(v-u)/a + (1/2)a(v-u)<sup>2</sup>/a<sup>2</sup>

or

s = (uv - u<sup>2</sup>)/a + (1/2)(v<sup>2</sup> - 2uv + u<sup>2</sup>)/a

or

2as = 2uv - 2u<sup>2</sup> + v<sup>2</sup> - 2uv + u<sup>2</sup>

or

2as = v<sup>2</sup> - u<sup>2</sup>

Rearranging, we get

v<sup>2</sup> = u<sup>2</sup> + 2as,

the final equation.

Notice that we didn't do any guessing. We started off with basic definitions, and worked through.

This is how vitually all equations in physics are derived.

Pete
07-02-03, 12:31 AM
Dimensional analysis is another useful method for deriving formulas.

I remember an example of a formula involving viscosity which was initially derived experimentally by a lot of hard work, but can be theoretically derived quite easily using dimensional analysis.

errandir
07-02-03, 01:31 PM
Pete, just to comment on the dimensional analysis. We should probably say that this is a better way to see how reasonable a formula is than to derive it. For instance:

If you want displacement from acceleration and time, following the dimensional analysis would give you a*t^2 as well as the correct result. But since this one seems more straightforward, it my erroneously be concluded strictly on the basis of dimensional analysis.

I am of course using an example that doesn't really need physics to derive; the actual formula is based on pure mathematical definitions. A better example would be kinetic energy, but I don't feel like going through that, so let's just assume that we don't know anything about calculus, in which case we would have to go through some inductive reasoning to find the 0.5 factor, or we would have to try to fit it to a curve and find that curve to be a parabola.

Pete
07-02-03, 06:49 PM
I agree that a formula derived by dimensional analysis must be tested to determine any constants involved, and that dimensional analysis is used more often as a sanity check than as a derivation tool.