# Pitot tube model

EDIT: Background on my application

I am making a few measurement models for an extended Kalman filter (EKF). The EKF is meshing an INS and a GPS currently, but the GPS update rate does not update fast enough to keep the filter converged. So I am tasked with adding some other measurement systems so that I can increase it's performance. Due to it being a Kalman filter I need the models to be linearized. So currently I want to linearize the Pitot tube model for that purpose. I am hoping if I find a good solution, I can also use it for other models with the same issues. So far, all of the work will be done purely as a simulation.

I'm trying to get a linearized model for a pitot tube, but I've got an asymptote in the linearized equation that I don't know how to deal with. I have Bernoulli's equation

$$P_o = P + 1/2\rho v^2$$

solved for velocity.

$$v = \sqrt{\frac{2(P_o-P)}{\rho}}$$

I would like a linearized equation so that I can use it as a linear model.

I've taken the derivative with respect to $$\Delta P$$

$$\Delta P = P_o - P$$

so that

$$\frac{dv}{dt} = \frac{1}{\sqrt{2\rho \Delta P}}\delta \Delta P$$

as $$\Delta P$$ approaches 0 I get an asymptote that I wouldn't get in the nonlinear equation.

My questions are:

1) Is this the proper approach to linearizing this equation?

2) How do I deal with an asymptote, in what should be the physical realm with no disconnect?

• Could you elaborate on your application, sound’s interesting – rul30 Mar 13 at 19:22
• I am making a few measurement models for an extended Kalman filter (EKF). The EKF is meshing an INS and a GPS currently, but the GPS update rate does not update fast enough to keep the filter converged. So I am tasked with adding some other measurement systems so that I can increase it's performance. Due to it being a Kalman filter I need the models to be linearized. So currently I want to linearize the Pitot tube model for that purpose. I am hoping if I find a good solution, I can also use it for other models with the same issues. So far all of the work will be done purely as a simulation. – Boto Mar 13 at 21:22

I am not sure, I understand what exactly are you trying to achieve, but I can try to answer your questions.

1. Is this the proper approach to linearizing this equation? All depends on purpose, but one usually takes a tangent at the point of interest as a linear approximation around this point. Which is what you did though your notation looks bit strange to me (what is dt for example?).

Anyway, you wonder why this does not work around the point $$\Delta P = 0$$, right? Take a look at the graph of $$\sqrt{x}$$, the tangent is "vertical" (parallel to $$y$$ axis) at the zero. So it does not cover any range of pressures and can not represent (approximate) any other close values of pressure.

The idea of linearization starts with expansion of the function $$v(\Delta p)$$ in proximity of some point $$\Delta p_0$$ into series: $$v(\Delta p) = v(\Delta p_0 + \varepsilon) = v(\Delta p_0) + \left.{dv\over d\Delta p}\right|_{\Delta p_0}\cdot\varepsilon + \left.{d^2v\over d\Delta p^2}\right|_{\Delta p_0}\cdot\varepsilon^2 + \left.{d^3v\over d\Delta p^3}\right|_{\Delta p_0}\cdot\varepsilon^3 + \cdots$$ If (!) for your small enough $$\varepsilon$$ the terms with higher derivation are much smaller than first derivation, you can neglect them and use linear term only.

Note that this assumption does not stay for square root in proximity of zero. In that case the higher derivations have higher value and can not be neglected. You can simply see it looking at the mentioned graph. If you place tangent line to square root graph, the line will stand away.

Depending on particular use, you can be better with using secant instead of tangent. You will get non-zero error in proximity of your chosen reference point, but overall error in larger surroundings can be smaller.

2. How do I deal with an asymptote, in what should be the physical realm with no disconnect? You simply do not try to approximate square root function with tangent through the origin. Use another suitable line (secant) minimizing overall error if you need linear approximation close to the origin.

This has nothing to do with physics though, the world is not guaranteed to be linear, neither in small scales.

• Hello, would it be possible to explain the secant approximation? I've only ever dealt with using a tangent line to approximate a nonlinear equation. – Boto Mar 14 at 13:52
• Also for clarification I have test data that changes over time. I consider $\Delta P$ to be time variant with the data. So I take the derivative $\partial v/\partial \Delta P$ which is the first derivative. Then I multiply the error which is an error (in my mind) that is with respect to time. That is how I rationalize it. essentially the form I have is $dv/dt = \partial v/\partial \Delta P * \partial \Delta P/ \partial t$ – Boto Mar 14 at 13:59