I have a friend who continously claims whilst were hiking that the reason she cant keep up is because she has short legs? Is there a calculation to determine who is using the most energy? one for the physics buffs I'm sure.
Person 1, 5ft2ins, 132 lbs
Person 2, 6ft2ins, 187lbs
Lets say they are walking at 4mph
Supplying the equation would be great.
This isn't really a question of energy (your friend can always eat more food to counter that), but the amount of work the leg muscles do.
My explanation would be roughly as follows: If we assume you and your friend are of similar build and body shape, your friend's legs will do about as much work in a stride as yours do, relative to their mass, if you both make strides at the same rate. However, your friend's legs are shorter, so to make up for it she has to make more strides per minute - which increases the workload by a factor of r^3 (one factor for the number of strides, two for the increased speed of each stride), where r is the ratio of leg lengths. In real life the body compensates somewhat, in that short people will end up with slightly larger leg muscles (relative to their total body mass) because of all this work. Still, even a small difference in leg length can make a big difference in how much effort is involved in walking/running.
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Originally posted by AcolyteThis surely can't be correct otherwise champion marathon runners would all be really tall. I think walking / running seems easier the lighter the non-muscle parts of your body are, which means it is easier for people with relatively puny skeletons rather than people with thick heavy bones. This does seem to be borne out by marathon runners, who appear mostly to have a fairly light build.
This isn't really a question of energy (your friend can always eat more food to counter that), but the amount of work the leg muscles do.
My explanation would be roughly as follows: If we assume you and your friend are of similar build ...[text shortened]... big difference in how much effort is involved in walking/running.