Assuming we have 2 identical cars. What is the total energy consumed by the two cars to start and accelerate up to a specific speed versus the total energy for one of them to accelerate up to a specific speed having the other tied to it? Or the fuel consumed for two vehicles to run on a road at 50mph and the fuel consumed for one vehicle tied to the other. I basically don't want an exact Joules number or Fuel litres. I need to know the difference between each of the cases. Which is most economical and by how much?
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2$\begingroup$ There are far too many unknown variables in this problem to provide any kind of answer. Consider editing it to make it more specific. $\endgroup$– grfrazeeJul 29, 2015 at 16:46
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$\begingroup$ what else could I write? I basically don't want an exact Joules number or Fuel litres. I need to know the difference between each of the cases. Which is most economical and by how much? $\endgroup$– sloupiocJul 29, 2015 at 17:02
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2$\begingroup$ Have a look at the Help Center. See here as well for some help as to what constitutes a good question. SE is more set up to answer specific, limited questions, not to answer general ponderings about a subject. $\endgroup$– grfrazeeJul 29, 2015 at 17:33
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1$\begingroup$ Also, if you hover your mouse over the "Thumbs Down" button, you will see that it says "This question does not show any research effort; it is unclear or not useful." Your question shows no research into your topic to give anyone a starting point, and your input parameters are unclear, or do not given enough input to begin solving the problem. (As an aside, you're mixing SI and Imperial units, which is generally a no-no). $\endgroup$– grfrazeeJul 29, 2015 at 17:38
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$\begingroup$ Related: Does having 2 engines together increase anything? $\endgroup$– AirJul 29, 2015 at 19:57
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1 Answer
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In your scenarios, there are two places that the energy ends up: as kinetic energy in the vehicles themselves (0.5 × mass × velocity squared) and as losses (rolling resistance, air resistance, etc.).
There are two possibilities regarding your two scenarios.
The two scenarios occur in the same amount of time. This means that the total energy is the same in the two scenarios, and that in the two-engine scenario, the engines are operating at less than half their capacity. In this case, the energy consumed would be based only on the relative efficiency of the engines at the two different power levels.
The single-engine scenario takes more time to reach the final speed than the two-engine scenario. In this case, while the final kinetic energy is the same, you have more losses while getting there, and therefore more energy input will be required, regardless of the engine efficiency numbers.
answered Jul 29, 2015 at 17:46
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$\begingroup$ your analysis starts well but I am not sure about the conclusion. Indeed in both cases (single engine dragging two vehicles VS two engines dragging each vehicle) the kinetic energy acquired is the same. So we should focus on the losses. At the moment I really cannot say if in one or the other case there are more losses. I have a "feeling" that if one engine drags two vehicles will be more efficient than having two engines dragging each vehicle. More moving parts, more friction. I am not sure, it doesn't seem clear enough. Any other idea?? $\endgroup$– sloupiocJul 29, 2015 at 18:50
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$\begingroup$ I tend to agree with @David Tweed, one addition to Scenario 2 is for both cars to reach the same speed in the same time as 2 cars operating there engines independently (Scenario 1), the one engine powering both cars would need to work much harder (if it was capable of doing so) & use more fuel $\endgroup$– FredJul 30, 2015 at 0:59
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$\begingroup$ yes, it would need to work harder, but usually engines are more efficient with increased torque/rpm. Also it is not the same to have a 2 ton car with having 1 ton car and another 1 ton car tied together. The rolling resistance increases, but not proportionally I think. $\endgroup$– sloupiocJul 30, 2015 at 12:26
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1$\begingroup$ Where did the 2-ton car come from? Until now, we have been discussing the "2 identical cars" that you posited originally. $\endgroup$ Jul 30, 2015 at 13:39