humerus

Question 1

humerus ap

Answer 1

• supine hand
• epicondyles parallel with ir
• both epicondyles seen in profile
• greater tubercle in profile

Question 2

humerus lateral

Answer 2

• hand in hips
• true lateral confirmed by superimposed epicondyles
• lesser tubercle in profile

Question 3

transthoracic lawrence method

Answer 3

• raise unaffected arm and place over the top of the head
• cr perpendicular surgical neck
• 10 - 15 cephalad if patient cant
• lateral view of the proximal 2/3 of the humerus

Question 4

shoulder ap neutral rotation

Answer 4

• palm of hand against thigh/hip
• IR 2 inches above shoulder
• cr perpendicular 1 inch inferior ro the coracoid process
• epicondyle 45 to ir
• oblique humerus
• greater tubercle partially superimposed

Question 5

shoulder ap internal rotation

Answer 5

• hand against thigh/hip
• epicondyles r perpendicular to the casette
• lesser tubercle in profile medially
• lateral view of humerus

Question 6

shoulder ap external rotation

Answer 6

• epicondyles are parallel
• greater tubercle in profile on the lateral aspect
• hand supine
• ap humerus

Question 7

inferosuperior axial lawrence method

Answer 7

• patient supine
• abduct arm of the affected side 90
• humerus in external rotation
• 15°-30° if abduction of arm is less than 90°
• lesser tubercle in profile directed
  anteriorly

Question 8

inferosuperior axial rafert modification

Answer 8

• abduct arm of the affected side 90°
• supine hand in exagerrated external rot
• 15° if di kaya 90
• lesser tubercle in profile
  directed anteriorly
• demonstrate hill-sachs defect

Question 9

inferosuperior axial west point method

Answer 9

• patient prone
• 25° anteriorly from the
  horizontal and 25° medially
  -

Question 10

inferosuperior axial clements modification

Answer 10

• patient in lateral recumbed
• abduct the affected arm 90° and point it toward the ceiling
• 5-15 if di kaya 90
• Lesser tubercle in profile

Question 11

scapular y PA oblique

Answer 11

• rotate patient’s body so that the midcoronal plane forms an angle of
  45°-60° to the IR.
• CR perpendicular to the scapulohumeral joint
• oblique image of the shoulder
• true lateral view of the scapula, proximal humerus

Question 12

glenoid cavity ap oblique grashey method

Answer 12

• rotate the body 35°-45°
  toward the affected side
• arm slightly with arm in neutral
  position
• CR perpendicular to the glenoid cavity at a point 2 inches (5cm) inferior to the superolateral border
  of the shoulder
• glenoid cavity in profile without superimposition of the humeral head

Question 13

glenoid cavity ap oblique apple method

Answer 13

• rotate the body 35°-45°
  toward the affected side
• abduct the arm 90° from the midline of the body holding a 1 pound weight on the affected side
• CR perpendicular to level
  of the coracoid process
• glenoid cavity in profile
• similar to grashey except for the use of the 1 pound weight

Question 14

glenoid cavity ap axial oblique garth method

Answer 14

• rotate the body 45°
  toward the affected side
• CR 45° caudad through
  the scapulohumeral joint
• acute shoulder trauma
• posterior scapulohumeral dislocations
• Anterior dislocation-Humerus
  projected inferiorly
• Posterior dislocationHumerus projected superiorly.

Question 15

tangential projection neer method

Answer 15

• unaffected side away from the IR 45°-60°
• CR 10°-15° caudad entering the superior aspect of the humeral head
• shoulder impingement/frozen suprespinatus oulet/coracoacromial

Question 16

scapular y PA oblique

Answer 16

• 45 - 60
• perpendicular
• shoulder dislocation

Question 17

stryker notch method

Answer 17

• flex arm slightly beyond 90
• 10 cephalad entering the coracoid process
• posterosuperior and posterolateral areas of the humeral head

Question 18

fisk modification

Answer 18

• intertubercular/tangential proj
• bicipital groove/depression greater tubercle and lesser tubercle

Question 19

acromoclavicular articulations ap pearson method

Answer 19

• patient upright
• SID 72 INCHES, include both joints,reduces distortion of the joint
• upright without weights and with weights (5-8lb) each wrist
• demonstrate AC joint disclocation, separation and function of the joints

Question 20

acromoclavicular articulations ap axial alexander method

Answer 20

• patient upright
• 15° cephalic to the coracoid process
• AC Joint projected slightly
  superiorly compared with
  an AP projection

Question 21

pa axial oblique alexander method rao/lao

Answer 21

• rotate the body 45-60°
  from the IR
• pull arm across the chest to draw the scapula laterally and forward
• CR 15 caudad to the AC joint
• AC joints in profile
• scapula and AC joint in the lateral position

Question 22

Blackett-Healy Method AP

Answer 22

• patient supine
• affected arm alongside thebody and rotate internally
• flex elbow
• pronate hand
• CR perpendicular to shoulder joint, entering coracoid process
• insertion of subscapularis at the lestter tubercle

Question 23

Blackett-Healy Method PA

Answer 23

• Pt prone
• IR under shoulder and center 1” below the coracoid process
• Arm in external rotation
• CR directly perpendicular
• Tangential image of
  insertion of the teres
  minor
• Lesser tubercle

Question 24

Zanca View AP Axial

Answer 24

• patient upright
• CR 10 cephalic
• ACJ free from superimposition
• Aid in assessment of
  distal osteophytes

Question 25

Serendipity View/Rockwood view

Answer 25

• patient supine/erect
• CR 40 cephalad SC joints
• Sternoclavicular joints
• Medial 1/3 of clavicles for fractures

Question 26

Hobbs Modification PA Transaxillary

Answer 26

• Arm raised superiorly as much as the patient can tolerate
• CR Perpendicular to axilla and
  humeral head to pass through the glenohumeral joint
• Lateral view of proximal humerus in relationship to Glenohumeral
  articulation
• Coracoid process of scapula
• Bursitis, Shoulder impingement, Osteoporosis, Osteoarthritis, Tendonitis