Mechanics

  • Objective
    • Moving the transducer where we know it’ll be
    • Having a strong probe body
  • Requirements
    • High speed rotation
    • Waterproofing the casing

Frame rate related to image definition

We calculate in this section the frame and the data quantity that we will need to send with our image specification(CAPTech section Image impact on tech - First level) and compare it to the one of the phd of Philippe Levesque (for the data quantity), cf. information given on CAPTech, section benchmark.

Here we consider that the speed of sound c is equal to 1540 m.s^{-1}.

For our purpose we have:

  • measurement width: d from 50 to 200 mm
  • sector angle: s = \pi/3
  • lateral resolution: r_l = 1 mm
  • number of lines: N_l = sR (R is the maximum with) = 210

In the phd of Philippe Levesque, we have:

  • measurement width d = 150 mm
  • sector angle: s = \pi/2
  • number of lines: N_l = 240
  • frame rate: f_r = 15 fps

Frame rate vs. number of lines

Depending on the number of lines in an image, it is simple to determine the maximum frame rate accessible to a probe. This frame rate depend if we use a sweep mechanism of a rotary mechanism.

Sweep mechanism

When we use a sweep mechanism, the image is done on the whole rotation of the motor, the maximum frame rate is given by:

f_r = \dfrac{c}{d_{max}N_l} = 36 fps. (1)

At contrary if we consider we want a frame rate at 20 fps, we can determine the maximum number of line at a given frame rate:

N_l = \dfrac{c}{d_{max}F_r} = 385 lines. (2)
Rotary mechanism

With this kind of mechanism, the transducer make a whole turn when the motor do also a turn, so it can be used only during the given sector angle. So with only one transducer, the maximum frame rate is:

f_r = \dfrac{c}{2d_{max}N_l}\dfrac{s}{2\pi} = 3 fps. (3)

So for our purpose, by using only one transducer, we must use a sweep mechanism to access a convenient frame rate.

Interpolating the position of the transducer

Theoretical formulation

With our mechanic setup, we can interpolate the position of the transducer. Indeed, the axis of the shelf of the transducer (which give straitly the angle of the transducer) is mathematically define par the intersection of two plans. This axis is located in the plan P_2 and the angle of the shelf is given by the plan p_3:

Vue cote.png

On this side view the axis is parallel to the axis y. \theta and r depend on the setup, we have \theta=\dfrac{\pi}{3} and r is undefined but it is not important (it is a constant that have no impact on the angle of the axis as we will see later). On our setup, the plan P_3 rotate around the axis z with an angle \phi as we can see above:

Vue dessus.png

The first image then correspond to \phi=0. For each \theta, we can make a variable change, the local frame (coordinate system) of the plan P_3 is then (X,Y,Z). In this local frame, we the plan is always define as in the first figure, so it can be express as:

P_3=\left(0;X,Y,-\left(x+r\right)tan\left(\theta\right)\right).

With the variable change, we have:

X=x\cos\left(\phi\right)-y\sin\left(\phi\right),
Y=y\cos\left(\phi\right)-x\sin\left(\phi\right).

So in its local frame, the plan P_3 can be expressed with the general coordinate as:

P_3=\left(0;x\cos\left(\phi\right)+y\sin\left(\phi\right),y\cos\left(\phi\right)-x\sin\left(\phi\right),\left(x\cos\left(\phi\right)+y\sin\left(\phi\right)+r\right)\tan\left(\theta\right)\right).

Then we have the intersection between P_2 and P_3:

P_3 \In P_2 = \left(0;y\sin\left(\phi\right),y\cos\left(\phi\right),-\left(y\sin\left(\phi\right)+r\right)\tan\left(\theta\right)\right).

Such as Z=z, the equation of the position of the axis is given by (center on the rotation axis: r=0):

z=-y\sin\left(\phi\right)\tan\left(\theta\right).

We search a en equation of the position of the axis on the form:

z=-y\tan\left(\psy\right),

where \psi is the angle of the axis on the global coordinate. So relation between \phi and \psy is:

\psi=\atan\left[\sin\left(\phi\right)\tan\left(\theta\right)\right],
\phi=\asin\left[\dfrac{\tan\left(\psi\right)}{\tan\left(\theta\right)}\right].

Time between two pulses

In order to form the image, we can choose that the angle between two pulses is constant or that the distance \Delta z between two pulse (on a plan normal to the axis) is constant. By knowing the angle difference \Delta \phi (of the motor position in radian) between two pulse, we can access to the time between two pulse at a given X fps :

\delta t=\dfrac{\Delta \phi}{pi*X}.

If we choose to have a constant \delta \psi so the shots will look like:

Regular pulse.png

and the time between two pulses is :

Dt dpsi.png

with this method, the minimal time between two pulses is 129 us at 15 fps and the maximal is 6969 us. The minimal time is less than the time of an acquisition, so we have to have a smaller fps or to change the interpolation method.

If we now choose to have a constant \Delta z, the shots will be:

Psi benoit.png

and the time between two pulses is:

Dt dz.png

with this method, the minimal time between two pulses is 212 us at 15 fps and the maximal is 3003 us.

Interesting pages

Bibliography

Couple of patents:

Subcategories

This category has only the following subcategory.

Media in category "Mechanics"

The following 12 files are in this category, out of 12 total.

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