Mandibular endosseous implant interventions 341

shafts are not tapered and have parallel walls with sharp line angles that add to their retentive qualities, the splint will be difficult to remove and could dislodge the implants.

The final fixed partial denture was tried in position, ground into proper occlusion, and a final intraoral periapical x-ray taken (Fig. 9-20). The bridge was then cemented into place with one of the permanent type cements (Fig. 9-21), thus completing the implant intervention procedure (Fig. 9-22).

Case 2

Unilateral posterior free-end saddle restoration using spiral-posts and a temporary splint

The spiral-shaft implant of Raphael Chercheve is a post type implant (Fig. 9-23) . This means that it can be used only where there is a fairly substantial

amount of alveolar bone. Only 4 to 5 mm. of the shaft should extend from the fibromucosal tissue. The length of this implant, however, can be varied by cutting off several spirals. Thus the spiraled portion can be set deeply enough in bone to allow a bony shelf to grow over it, making exfoliation extremely difficult.

The insertion technique for the spiral-shaft implant is similar to that used for inserting the vent-plant, except that special taps must be used to trephine a hole exactly the size of the implant. This step is eliminated when using the vent-plant, which is self-tapping.

In the following case, two Chercheve spiral-post implants and one natural tooth are used to support a five-unit fixed partial denture.

First, intraoral periapical radiographs were taken

Fig. 9-23. A, The spiral-shaft double-helical implants. B, A good portion of the shaft should extend into the bone below the alveolar crest (A') so that the eventual invagination of the epithelial tissue (B') would not extend to the open spirals. C, Only about 4 to 5 mm. of the shaft (A-B) should extend out of the fibromucosal tissue. The rest of the implant (B-C) should be buried in bone. The masticatory forces (FM) are applied to the extraosseous portion of the implant while resistance to this force (FR) is opposed by the intraosseous portion (BC). Therefore: FM x AB = FR x BC.