Difference between revisions of "Store:ACvericale"

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(Created page with "==Descrizione delle misure lineari ed angolari== ===Rappresentazione scalare dei tracciati condilari=== '''Descrizione delle distanze e delle direzioni''' Di seguito sono riportate le distanze calcolate tra i punti rispetto al punto di partenza (punto 1) considerato il unto di riferimento essendo la mandibola in una posizione di Massima Intercuspidazione e le relative direzioni nello spazio, utilizzando le coordinate corrette per gli assi <math>X</math> (antero-poster...")
 
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'''Punto 2L'''
'''Punto 2L'''


Coordinate: (59.0, -92.3) Calcolo della distanza rispetto a 1L: <math>
Coordinate: (59.0, -92.3) Calcolo della distanza 'd'rispetto a 1L: <math>
\text{distanza} = \sqrt{(59.0 - 58.3)^2 + (-92.3 + 50.9)^2} = \sqrt{(0.7)^2 + (-41.4)^2} \approx \sqrt{0.49 + 1714.56} \approx \sqrt{1715.05} \approx 41.42 \, \text{pixel}
d= \sqrt{(59.0 - 58.3)^2 + (-92.3 + 50.9)^2} = \sqrt{(0.7)^2 + (-41.4)^2} \approx \sqrt{0.49 + 1714.56} \approx \sqrt{1715.05} \approx 41.42 \, \text{pixel}
</math> Distanza in millimetri:  <math>  
</math> Distanza in millimetri:  <math>  
41.42 \, \text{pixel} \times 0.1 \, \text{mm/pixel} = 4.14 \, \text{mm}  
41.42 \, \text{pixel} \times 0.1 \, \text{mm/pixel} = 4.14 \, \text{mm}  
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<math>
<math>
\text{distanza} = \sqrt{(46.3 - 58.3)^2 + (-169.5 + 50.9)^2} = \sqrt{(-12.0)^2 + (-118.6)^2} \approx \sqrt{144.0 + 14065.96} \approx \sqrt{14209.96} \approx 119.2 \, \text{pixel}
d= \sqrt{(46.3 - 58.3)^2 + (-169.5 + 50.9)^2} = \sqrt{(-12.0)^2 + (-118.6)^2} \approx \sqrt{144.0 + 14065.96} \approx \sqrt{14209.96} \approx 119.2 \, \text{pixel}
</math>
</math>


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Coordinate: (44.1, -207.7)  Calcolo della distanza rispetto a 1L: <math>
Coordinate: (44.1, -207.7)  Calcolo della distanza rispetto a 1L: <math>
\text{distanza} = \sqrt{(44.1 - 58.3)^2 + (-207.7 + 50.9)^2} = \sqrt{(-14.2)^2 + (-156.8)^2} \approx \sqrt{201.64 + 24596.84} \approx \sqrt{24798.48} \approx 157.5 \, \text{pixel}
d = \sqrt{(44.1 - 58.3)^2 + (-207.7 + 50.9)^2} = \sqrt{(-14.2)^2 + (-156.8)^2} \approx \sqrt{201.64 + 24596.84} \approx \sqrt{24798.48} \approx 157.5 \, \text{pixel}
</math>Distanza in millimetri: <math>  
</math>Distanza in millimetri: <math>  
157.5 \, \text{pixel} \times 0.1 \, \text{mm/pixel} = 15.75 \, \text{mm}
157.5 \, \text{pixel} \times 0.1 \, \text{mm/pixel} = 15.75 \, \text{mm}
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<math>
<math>
\text{distanza} = \sqrt{(38.4 - 58.3)^2 + (-136.2 + 50.9)^2} = \sqrt{(-19.9)^2 + (-85.3)^2} \approx \sqrt{396.01 + 7276.09} \approx \sqrt{7672.1} \approx 87.6 \, \text{pixel}
d= \sqrt{(38.4 - 58.3)^2 + (-136.2 + 50.9)^2} = \sqrt{(-19.9)^2 + (-85.3)^2} \approx \sqrt{396.01 + 7276.09} \approx \sqrt{7672.1} \approx 87.6 \, \text{pixel}
</math>
</math>


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<math>
<math>
\text{distanza} = \sqrt{(36.4 - 58.3)^2 + (-48.2 + 50.9)^2} = \sqrt{(-21.9)^2 + (2.7)^2} \approx \sqrt{479.61 + 7.29} \approx \sqrt{486.9} \approx 22.1 \, \text{pixel}
d = \sqrt{(36.4 - 58.3)^2 + (-48.2 + 50.9)^2} = \sqrt{(-21.9)^2 + (2.7)^2} \approx \sqrt{479.61 + 7.29} \approx \sqrt{486.9} \approx 22.1 \, \text{pixel}
</math>
</math>


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<math>
<math>
\text{distanza} = \sqrt{(44.0 - 58.3)^2 + (-34.9 + 50.9)^2} = \sqrt{(-14.3)^2 + (16.0)^2} \approx \sqrt{204.49 + 256.0} \approx \sqrt{460.49} \approx 21.5 \, \text{pixel}
d = \sqrt{(44.0 - 58.3)^2 + (-34.9 + 50.9)^2} = \sqrt{(-14.3)^2 + (16.0)^2} \approx \sqrt{204.49 + 256.0} \approx \sqrt{460.49} \approx 21.5 \, \text{pixel}
</math>
</math>


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<math>
<math>
\text{distanza} = \sqrt{(52.9 - 58.3)^2 + (-48.0 + 50.9)^2} = \sqrt{(-5.4)^2 + (2.9)^2} \approx \sqrt{29.16 + 8.41} \approx \sqrt{37.57} \approx 6.13 \, \text{pixel}
d= \sqrt{(52.9 - 58.3)^2 + (-48.0 + 50.9)^2} = \sqrt{(-5.4)^2 + (2.9)^2} \approx \sqrt{29.16 + 8.41} \approx \sqrt{37.57} \approx 6.13 \, \text{pixel}
</math>
</math>


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