We're going to call the horizontal edge e, the front two edges a and b, and the back two legs c and d. Also, we see that the numbers 11, 10 and 9 must all be in feet, as they're too big to be anywhere else.

Let's try S=14 first. We may as well take a=3, as the legs are all interchangeable at this stage.

Now if b=4, then we have c=5 (as the back legs are interchangeable). Also, we must have d and the back foot as 6 and 8(in some order), and we soon see that this isn't possible (as we'd also need e plus the top left hip to add up to 7). If b isn't 4, then we can assume that c=4. But similarly we find that we can't have b=9 or b=8 or b=7. So S isn't 14.

Let's try S=16. But then we need four pairs of numbers that sum to 16 (one for each leg). We try: (11,5), (10,6), (9,7), (8,?)and we can't repeat the 8 so that isn't possible.

So let's try S=15. This time, we may as well take a=4. There are only 4 pairs that add up to 15: (11,4), (10,5), (9,6) and (8,7). Call the number in the top left hip f.

If b=5, then we have c=6 and e+f=6, which isn't possible (as we don't have the right numbers left).

So we can take c=5. If b=8, have e+f=3 and then we'll need f+g=4, and this isn't possible. If b=7, have e+f=4 and we also have f+g=4 and this isn't possible either. Also we can't have d=8, as this is too big. So the only possibility is the one shown below (and this does indeed work).

(When I say the only possibility, of course the front legs could be switched, or the back legs, or the front and back swapped, but it's really the same caterpillar.)