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 7 added 30 characters in body edited Apr 19 '16 at 19:05 flawr 1,67766 silver badges1515 bronze badges Again looking at the graph above, we can conclude that if the weights are close enough, I suggest the rule of thumb the relative difference in velocity is going to be about half the relative difference in mass.the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that: Again looking at the graph above, we can conclude that if the weights are close enough, the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that: Again looking at the graph above, we can conclude that if the weights are close enough, I suggest the rule of thumb the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that: 6 added 503 characters in body edited Apr 19 '16 at 18:58 flawr 1,67766 silver badges1515 bronze badges Again looking at the graph above, we can conclude that if the weights are close enough, the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that: If you have a weight difference of 10% (as long as the weights are close enough, it does not matter which one you consider as 100%) then the difference in velocity is going to be about 5%. Again checking with the "exact" formula (considering `m1 = 100%`)``````v2 = sqrt(m1/m2) * v1 = sqrt(1.1) * v2 = 1.048 * v1 so that is about 5% `````` Alternatively if we consider `m2 = 100%```````v2 = sqrt(m1/m2) * v1 = sqrt(0.9) * v2 = 0.948 * v1 so that is about 5% too. `````` Again looking at the graph above, we can conclude that if the weights are close enough, the relative difference in velocity is going to be about half the relative difference in mass. Again looking at the graph above, we can conclude that if the weights are close enough, the relative difference in velocity is going to be about half the relative difference in mass. What do I mean by that: If you have a weight difference of 10% (as long as the weights are close enough, it does not matter which one you consider as 100%) then the difference in velocity is going to be about 5%. Again checking with the "exact" formula (considering `m1 = 100%`)``````v2 = sqrt(m1/m2) * v1 = sqrt(1.1) * v2 = 1.048 * v1 so that is about 5% `````` Alternatively if we consider `m2 = 100%```````v2 = sqrt(m1/m2) * v1 = sqrt(0.9) * v2 = 0.948 * v1 so that is about 5% too. `````` 5 added 503 characters in body edited Apr 19 '16 at 18:52 flawr 1,67766 silver badges1515 bronze badges I do not know any rule of thumbs but I do have some knowledge in physics and math, so lets see what we can do with that=) Let's assume the energy that is put into an arrow is independent of the weight of the arrow. If we would want to model this too, it would be highly dependent of construction of the bow, and not useful for any rule of thumb. And as long as the weight of the two arrows we want to consider is sufficiently similar, this assumption will still provide good enough results. Let's assume the energy that is put into an arrow is independent of the weight of the arrow. If we would want to model this too, it would be highly dependent of construction of the bow, and not useful for any rule of thumb. And as long as the weight of the two arrows we want to consider is sufficiently similar, this assumption will still provide good enough results. I do not know any rule of thumbs but I do have some knowledge in physics and math, so lets see what we can do with that=) Let's assume the energy that is put into an arrow is independent of the weight of the arrow. If we would want to model this too, it would be highly dependent of construction of the bow, and not useful for any rule of thumb. And as long as the weight of the two arrows we want to consider is sufficiently similar, this assumption will still provide good enough results. 4 added 503 characters in body edited Apr 19 '16 at 18:46 flawr 1,67766 silver badges1515 bronze badges 3 added 338 characters in body edited Apr 19 '16 at 18:31 flawr 1,67766 silver badges1515 bronze badges 2 added 338 characters in body edited Apr 19 '16 at 18:25 flawr 1,67766 silver badges1515 bronze badges 1 answered Apr 19 '16 at 18:20 flawr 1,67766 silver badges1515 bronze badges