You must first make sense of the orientation. If you are travalling horizontal, will your inclinomitor measure 90º or 0º.

Really, just up and down? Once? Or is this Down and then Up?The shape of the compressed beam can be approximated as a half sine wave, somewhat like this, ")" in real life.

Let's see if I have this. We are essentially just going down, but compared to a straight drop (I'll call it a "pole"), we wander off for a while and then return. Is the start and stop abrupt or do we get a mitigation at the endpoints?Thus, the inclinometer outputs the slope in degrees along the beam.

You certainly should modify this to , so that the amplitude (maximum horizontal displacement from vertical) is 'a' and the length of the beam is half the period of the sine function. If the length is 'L', then or Finally, 'c' is where the nose starts off the track, since I presume it is not actually in contact with the track. Units are whatever you measure in. Feet? Meters? x is vertical displacement down the pole. s(x) is horizontal displacement away from the pole.distance vs amplitude: sin(x) from x 0 to pi.

. Units here are, x is still vertical displacement down the pole.then taking the derivative to get the slope: cos(x)

(slope in what units? unit-less? wouldn't that be radians?)

v(x) is the speed at which the horizontal displacement is changing. This will depend on your meansurements. feet/sec, meter/ms, whatever.

This is why you hire professionals. You are hung up on concepts that are either not significant or don't mean what you think they mean, or in this case, don't mean anything at all. A radian is not a unit. It is a fake unit so people feel better about dropping degrees. It's just a number.to get radians: arctan(cos(x))

and degrees: arctan(cos(x)*(180/pi))

Your data should present an excellent set of observed data for determining a good approximate model for v(x). Again, this may be a job for a professional. When was the last time you used a Moore-Penrose inverse? With 1000 points and only three parameters, you should get an excellent model. On the other hand, if you already know the two paremeters, a, b, and c, why not go the other way and calculate the exact model? I know the answer, of course. The sine wave is only your visual interpretation of the structure. If it were the top of a parabola or an inverted catenary, would your eye know the difference?I have about 1000 data points in excel that are the slope along the rail in degrees vs distance from the start. I want to numerically integrate this data to get a graph of the amplitude of the bow along the rail vs distance.

Should I convert my degree slope data to radians, then take the tangent of that to get the actual slope "m"? Then integrate that data? Or integrate the radian data? What would be the physical significance (units?) of both methods?

If this is a project that will risk someone's life, you will want as good a model as can be created. For this, the data should be examined for appropriateness and the actual structure discerned.

I'm always up to a long-winded response in return. What say you? Did we get anywhere?Sorry if that was long winded,