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| What kind of BBs do you use? | |
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| Tweet Topic Started: Nov 13 2011, 09:26 PM (499 Views) | |
| surfer | Nov 14 2011, 03:10 PM Post #11 |
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Ah another classic lecture here from Nova. So what about taking the derivative of that? |
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| Nova | Nov 14 2011, 03:17 PM Post #12 |
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Polar98 CO
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The derivative can verify the results only after some data has been collected. Well if you made f(x) the .20g and g(x) the .25g, you'd notice two different curves. d/dx (KE) = (1/2) dm/dt(v)+2dv/dt(m) == v(dm/dt)+m(dv/dt) Alright, so here's the derivative with respect to time. This is interesting because mass and velocity are now equally weighted. In fact, this almost looks like the momentum equation, but we'll discuss that another day. In this equation using the initial m and v for each would retain the same value and be linear. This would work if you were in a vaccume. In real life we have air resistance which gives us the two curves f(x) and g(x). Due to this, the curve f(x) will have a more negative derivative than g(x). In f(x), v is decreasing rapidly while retaining a smaller amount of energy as mass. This in effect verifies the thesis previously discussed. And for the record, the second derivative doesn't really reveal anything except for the similarities of the properties. d2y/dx^2 = v(m'')+(dm/dt)v' + m(v'')+(dv/dt)m'' The second derivative should always be positive for this meaning that the original function is always concave up. The negative first derivative signifies a decrease in the value of the function (negative slope/tangent). So we end up with a decay curve, which is what happens in this case. All this really shows is the general similarities of properties. Edited by Nova, Nov 14 2011, 03:40 PM.
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| surfer | Nov 14 2011, 03:27 PM Post #13 |
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5000+ posts
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Culd you post a graph for f(x) and g(x) for refrence? Just kidding. I know exactly what you mean.
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| Nova | Nov 14 2011, 03:38 PM Post #14 |
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Polar98 CO
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If the function surfer posted is f(x), g(x) would have the same start point but a smoother slope and hit the x axis later than f(x). As you can see that function is concave up and decreasing. The graph of the derivative would be similar because it is a property of exponential growth and decay to have a similar derivative graph to the original function. In fact the derivative of the function e^x is e^x. Same goes for the second derivative. But this now brings to light something worth considering. Q: What if we took the integral of the original equation. I don't even know if implicit integration is possible without multi-variable calculus, but if there wasn't multiple variables in there, we could do it. Well, thanks for this discussion. And enjoy. OH WOW: It might be possible to calculate the point at which it is worth it to use a .25g over a .23g or something. Since we don't know what air resistance is (it is dynamic in nature) we can't find exactly where this is yet. But rest assured it is possible to do in a controlled environment. This would be found at the point in which g'(x) is less negative than f'(x). Edited by Nova, Nov 14 2011, 03:46 PM.
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| surfer | Nov 14 2011, 04:37 PM Post #15 |
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Excellent presentation. |
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| vegas | Nov 14 2011, 05:43 PM Post #16 |
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"The German"
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Good thing I remembered a thing or two about physics last year....I actually understood some of that. |
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Airsoft Drifter S.T.A.L.K.E.R. Airsoft Association Content Creator Regional Events attended: SOG-MD- Op Sunset Local Events: Polar98 events Op: Snowfall | |
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| surfer | Nov 14 2011, 06:03 PM Post #17 |
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I had physics in high school. But I had Calculus in my second year of college. Nova happened to use Calculus to justify his points instead of just algebra. It's actually better to use calc if you know how |
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