WEEK 5 - SAMPLE LAB EXAM QUESTION
- QUESTION:
Pteranodon had a wingspan of about 7 meters - twice that of the largest living bird. Pteranodon's wing was supported by one extremely elongated digit, as seen in the model.
Figure A shows a cross-section of a typical mammal bone (a camel tibia) and a cross section of the first bone of the long finger of Pteranodon
Calculate the k value of these two bones. Explain why the optimal k values might be very different for these two animals.
- ANSWER:
Camel tibia: k=approximately 0.56
Pteranodon: k=approximately 0.93
Pteranodon's wing bones are air filled, making them much lighter than the marrow filled bones of the camel. Light wing bones are advantageous to a large flying animal. For bones filled with marrow, k values of 0.63 are optimal, minimizing weight for a bone of a given strength. Air-filled bones have a much higher optimal k value, as weight of a hollow bone of a given strength continues to drop as the ration of internal to external diameter of the bone increases.
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