Thread: Real World Problem - Integrating Slope Data!

Hi,

I am a bit confused on the meaning of "slope" and the first and second derivatives.

I have an inclinometer that measures its angular orientation with respect to pure vertical, and it is traveling on a small cart up a beam that is compressed. The shape of the compressed beam can be approximated as a half sine wave, somewhat like this, ")" in real life. Thus, the inclinometer outputs the slope in degrees along the beam.

To approximate this I can construct a sine wave ie:

distance vs amplitude: sin(x) from x 0 to pi.

then taking the derivative to get the slope: cos(x)
(slope in what units? unit-less? wouldn't that be radians?)

to get radians: arctan(cos(x))

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

This gives the slope along the simulated rail in degrees. My question is this:
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?

This is kind of confusing because distance -> velocity -> acceleration when you are with respect to time, but I am plotting amplitude (distance) vs distance along the rail.

Sorry if that was long winded,

Thanks!

Originally Posted by stevenp17

I have an inclinometer that measures its angular orientation with respect to pure vertical,

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

The shape of the compressed beam can be approximated as a half sine wave, somewhat like this, ")" in real life.

Really, just up and down? Once? Or is this Down and then Up?

Thus, the inclinometer outputs the slope in degrees along the beam.

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?

distance vs amplitude: sin(x) from x 0 to pi.

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.

then taking the derivative to get the slope: cos(x)
(slope in what units? unit-less? wouldn't that be radians?)

. Units here are, x is still vertical displacement down the pole.
v(x) is the speed at which the horizontal displacement is changing. This will depend on your meansurements. feet/sec, meter/ms, whatever.

to get radians: arctan(cos(x))

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

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.

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?

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?

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.

Sorry if that was long winded,

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

Don't worry, were just doing this on a simple test rig, no ones life is in danger.





Those are the more advanced equations I've been using in a MATLAB simulation.

I don't see where velocity enters into this to get meters/s. My only data is vertical distance off of the ground in meters, and at those points I have a inclination reading, time independent. Perhaps we are getting caught up on the math, I only chose a half sine wave because it is a rough approximate of the curvature of a pinned-pinned beam under compression.

If I have the slope values in degrees vs the distance traveled, how do I integrate this to get position vs distance? My trouble lies in the question: what slope values to integrate, the pure ratio rise/run "m", radians, degrees? What are the units / meaningful representations of these integrations?

I am working with measured angular devations that are small, say +/- 2 degrees max, so these bowing amplitudes are rather small over a beam of say 2 meters.

I see. I was starting at the top of the pole and heading downward.

Let's see if we can get on the same page.

When you start out, you are nearly vertical with an upward orientation. Your inclinomitor is nearly zero. Right? If so, this is quite a blow to the sine function. Not to worry, though, since we don't really care. What is needed is a good model. It does not matter if it is some known trigonometric function.

Rough Example:

Sample every 10 cm of vertical displacement

Inclination(00 cm) = 0.00º ==> 0.000
Inclination(10 cm) = 1.00º ==> 0.017
Inclination(20 cm) = 1.50º ==> 0.026
Inclination(30 cm) = 1.80º ==> 0.031
Inclination(40 cm) = 1.95º ==> 0.034
Inclination(50 cm) = 2.00º ==> 0.035

Reconstruct original function from this 1st derivative information.

Extremely Rough (Right Rectangular)

Displacement(10) = 10*(0.017) = 0.170
Displacement(20) = 10*(0.017 + 0.026) = 0.430
Displacement(30) = 10*(0.017 + 0.026 + 0.031) = 0.740
Displacement(40) = 10*(0.017 + 0.026 + 0.031 + 0.034) = 1.080
Displacement(50) = 10*(0.017 + 0.026 + 0.031 + 0.034 + 0.035) = 1.430

Better (Trapezoid)

Displacement(10) = 10*(0.000 + 0.017)/2 = 0.085
Displacement(20) = 10*(0.000 + 2*0.017 + 0.026) = 0.300
Displacement(30) = 10*(0.000 + 2*0.017 + 2*0.026 + 0.031) = 0.585
Displacement(40) = 10*(0.000 + 2*0.017 + 2*0.026 + 2*0.031 + 0.034) = 0.910
Displacement(50) = 10*(0.000 + 2*0.017 + 2*0.026 + 2*0.031 + 2*0.034 + 0.035) = 1.255

There are many methods to perform simple integration. I would start with the trapezoid rule and see if that is sufficient for you. With 1000 points over a measly 2 metres, I suspect it may be close enough for the kinds of horizontal displacement you are talking about.

ALL of this has no reference at all to any known function. It merely defines what it is.