u still walk for a while in pain?
2006-08-28 14:14:27 · 10 answers · asked by Anonymous in Society & Culture ➔ Etiquette
Morbid question, are you planning on leaving someone stranded?
NO. That muscle is needed to hold the foot and leg firmly. anyone cut that way would have to fashion a crutch and limp along on the good leg.
2006-08-28 14:22:09 · answer #1 · answered by mindbender - seeker of truth 5 · 0⤊ 0⤋
Well at skating practice a girl got her calf cut in half but she just passed out cause she lost a lot of blood, i dont think she could walk, but its been 3 months and she can walk with a cane now, and soon they say she will be able to go back to skating.
2006-08-28 21:21:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
It would probably be painful and you'd buckle I mean you use your calf muscle to walk and stand.
2006-08-28 21:27:15 · answer #3 · answered by ♥Nina♥ 3 · 0⤊ 0⤋
answer to the question is this
no you wont walk a while, you wont buckle either, neither will you pass out immediately( none of these will happen immeiately)
you will be able to take a step or two however before you get unconsious cos of blood loss.
2006-09-01 17:51:50 · answer #4 · answered by baby blu 3 · 0⤊ 0⤋
I don't think you could keep walking because you can't cut the muscle without cutting the tendon.
2006-08-28 21:18:29 · answer #5 · answered by maigen_obx 7 · 0⤊ 0⤋
See "Monty Python and The Holy Grail" for what you seek. The "Black Knight" who guards the bridge holds your answer.
2006-08-31 11:10:26 · answer #6 · answered by YahooGuru2u 6 · 0⤊ 0⤋
I think you would be down and out for the count.
2006-08-28 22:09:34 · answer #7 · answered by turtle girl 7 · 0⤊ 0⤋
you would go down like a sack of potatoes.
2006-08-28 21:33:47 · answer #8 · answered by Steve K 2 · 0⤊ 0⤋
I would just hop on my good leg.
2006-08-28 21:20:08 · answer #9 · answered by Like Bugs Bunny on the moon... 2 · 0⤊ 0⤋
on one leg. you wouldnt be able to push off with your foot. you would be able to stand on one foot.
2006-08-29 15:53:19 · answer #10 · answered by doc2be 4 · 0⤊ 0⤋