Shoulder-details, projections, rotations, etc.
Order by
16 terms
Terms | Definitions |
|---|---|
 1 inch inferior to the coracoid process | For an AP External, Neutral and Internal Rotation the CR should be positioned _______. |
| 24 x 30 (or) 10 x 12 -Crosswise | Image Receptor Size for an AP Shoulder (External, Neutral and Internal rotation) is _____. |
| External Rotation | The epicondyles are parallel with the plane of the IR in which rotation? |
| Neutral Rotation | The epicondyles are about 45 degrees with the plaine of the IR in which rotation? |
| Internal | The epicondyles are perpendicular with the plane of the IR in which rotation? |
| External Rotation | In which rotation would you see: Humeral head in profile, Greater tubercle in profile on the lateral aspect of the humerus, Scapulohumeral joint visualized with slight overlap of humeral head on glenoid cavity, outline of lesser tubercle between the humeral head and greater tubercle. |
| Neutral Rotation | In which rotation would you see: Greater tubercle partially superimposing the humeral head, Humeral head in partial profile, slight overlap of the humeral head on the glenoid cavity |
| Internal Rotation | In which rotation would you see: Lesser tubercle in profile and pointing medially, outline of the greater tubercle superimposing the humeral head, Greater amount of humeral overlap of the glenoid cavity than in the external and neutral positions |
| Transthoracic Lateral Projection (Lawrence Method) | In which rotation would you see: A lateral image of the shoulder and proximal humerus is projected through the thorax. |
| Transthoracic Lateral Projection (Lawrence Method) | Image Receptor: 24 x 30 cm lengthwise |
| Transthoracic Lateral Projection (Lawrence Method) | For the what method will the CR will enter the mnidcoronal plane at the level of the sugical neck. |
| Transthoracic Lateral Projection (Lawrence Method) | For this projection you will see: Proximal humerus, Scapula, clavicle and humerus seen through the lung field, scapula superimposed over the thoracic spine, unaffected clavicle and humerus projected above the shoulder closest to the IR, humeral head in neutral rotation. |
| Inferorsuperior Axial Projection (Lawrence Method) | In which rotation would you see: Scapulohumeral joint with slight overlap, coracoid process, pointing anteriorly, lesser tubercle in profile and directed anteriorly, AC joint acromion and acromial end of clavicle projected through the humeral head |
| Inferorsuperior Axial Projection (Lawrence Method) | The CR for this method is directed horizontally through the axilla to the region of the AC articulation. The degree of medial angulation of the central ray depends on the degree of abduction of the arm. The degree of medial angulation is often between 15 and 30 degrees. (The greater the abduction the greater the angle) Ex: if the arm is at 90-degree angle, then the CR is at 25-30 degree |
| Inferorsuperior Axial Projection (Lawrence Method) | Image Receptor: 24 x 30 cm or 10 x 12 crosswise |
| Inferorsuperior Axial Projection (Lawrence Method) | In this rotation you would see: Scapulohumeral joint with slight overlap, coracoid process, pointing anteriorly, lesser tubercle in profile and directed anteriorly. AC joint, acromion and acromial end of clavicle projected through the humeral head |
First Time Here?
Welcome to Quizlet, a fun, free place to study. Try these flashcards, find others to study, or make your own.