Wikipedia:Reference desk/Archives/Mathematics/2012 February 7

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February 7

Trains allegedly haul 1 ton for nearly 500 miles on a gallon. How to calculate?

I saw on a commercial about a rail freight corporation that trains will haul 1 ton of cargo nearly 500 miles on a single gallon of fuel.

Would anyone please demonstrate how this statistic is calculated? Thank you. --70.179.174.101 (talk) 07:41, 7 February 2012 (UTC)

For one ton of cargo, you'd use:

(MILES TRAVELED)/(GALLONS OF FUEL USED)

However, I suspect that they didn't count the fuel needed to move the train itself:

(MILES TRAVELED)/(GALLONS OF FUEL USED BEYOND THAT FOR EMPTY TRAIN)

If so, then this statistic is rather useless. After all, almost all of the fuel used in a car or plane is also used to move the car or plane itself, and not it's cargo. StuRat (talk) 08:55, 7 February 2012 (UTC)

The formula used must be

(MILES TRAVELED)×(TONS OF CARGO)/(GALLONS OF FUEL USED)

Bo Jacoby (talk) 10:27, 7 February 2012 (UTC).

I would expect they mean miles per gallon for tons of cargo and the gallons is total over their whole operation so in fact it is quite a bit better than one might think. Trains have lower air resistance and 'tire' resistance, they go at an even speed for long distances and are quite good at recovering energy when braking, basically a diesel electric locomotive is a huge and extremely efficient Toyoto Prius. Dmcq (talk) 10:34, 7 February 2012 (UTC)

Diesel-electric trains certainly are efficient, but while the locomotive may employ dynamic braking, power will typically be dissipated in brake grid resistors, not stored in batteries as on a Prius. Electric trains connected to overhead lines or a third rail may return energy to the system during braking. (Freight trains usually try to avoid stop-and-go traffic.) -- ToE 01:24, 9 February 2012 (UTC)

We do have an article on Hybrid trains, but that are not currently in common use. -- ToE 01:39, 9 February 2012 (UTC)

Yes it looks like the ones with batteries are only being used in any great number for shunting rather than long haul and one would only get power returned for the others with overhead lines where it would just need an electric train anyway. Dmcq (talk) 02:05, 9 February 2012 (UTC)

Something else to consider is that trains only move cargo from one train depot to another. Unless your factory is at one depot and customer at another, you will therefore incur additional shipping costs for delivery. StuRat (talk) 10:45, 7 February 2012 (UTC)

We have an article Fuel efficiency in transportation which may be of some use.--Salix (talk): 17:43, 7 February 2012 (UTC)

Note that the 185.363 km/l (per short ton) quoted there is 436 mi/gal. -- ToE 01:39, 9 February 2012 (UTC)

Not too bad but not quite 500 but they must have had a reason for that, I wonder if having better locomotives in shunting yards could have made the difference. I'm not sure if I read it right but it looks to me that for passenger trains something like half the power is used in the passenger cars rather than for driving the train along! Dmcq (talk) 02:05, 9 February 2012 (UTC)

They might also be talking about movement on straight, level ground, with no accel or decel (so excluding starts and stops). StuRat (talk) 02:19, 9 February 2012 (UTC)

I'm pretty certain they mean delivered cargo and total fuel for the freight trains over the whole operation. Association of American Railroads: Freight Railroads Help Reduce Greenhouse Gas Emission in 2010 describes what they are doing. Dmcq (talk) 10:31, 9 February 2012 (UTC)

Angle bisector problem

Can someone help me how to do this? Triangle ABC has AB = 27, AC= 26, BC = 25. Let say I is the incenter point, where all three angle bisectors met. Find BI. Thanks!Pendragon5 (talk) 23:51, 7 February 2012 (UTC)

ABI is another triangle. Use the cosine rule to find the angles in ABC, divide the angles at points A and B by 2 to get the angles of this new triangle (ABI), then use the sine rule to find the length of BI. Widener (talk) 00:43, 8 February 2012 (UTC)

Ops forgot to say. No calculator is allowed.Pendragon5 (talk) 01:12, 8 February 2012 (UTC)

Is this homework? WP:DYOH RudolfRed (talk) 03:19, 8 February 2012 (UTC)

Nah?? If it is homework then i'm pretty sure i can do it by myself. Why would i not know how to do it unless i was not paying attention in class, which is not my type. If you wonder where this problem comes from then let me tell you it is part of the AMC 12 test, which i just took it today. And they are not easy. Don't try to advise me on whether or not i do my homework. Even if this is my homework, which it is not, so what? You're not helping me that's your choice. I have the right to ask.Pendragon5 (talk) 03:52, 8 February 2012 (UTC)

Chill out man. The policy of the reference desk is that we don't do people's homework for them. It's not obvious from your question that it's not homework, so RudolfRed asked. How can we know what "your type" is?

As for the question, drop the perpendiculars from the incenter, and you get 3 pairs of congruent right triangles. You can find the inradius from the side lengths with the formula in the article (it's √56). You can solve for the lengths of the remaining legs of the right triangles x,y,z by setting up the three equations x+y = 27, x+z = 26, y+z = 25 and find that y is 13. Then by Pythagorean theorem, BI is 15. Rckrone (talk) 06:02, 8 February 2012 (UTC)

If you don't know the formula for calculating the inradius, then you can also find it in the following way: The triangles ABI, AIC, and IBC all have height equal to the inradius r and base 27, 26, and 25 respectively. Therefore, their areas are 13.5r, 13r, and 12.5r respectively. The sum of these areas is equal to the area of the whole triangle ABC and is equal to 39r. From Heron's formula (which you should probably memorize) you also know that the area of ABC is ${\sqrt {39\cdot 12\cdot 13\cdot 14}}$ so we have $39r={\sqrt {39\cdot 12\cdot 13\cdot 14}}\implies r={\sqrt {\frac {12\cdot 13\cdot 14}{39}}}={\sqrt {\frac {12\cdot 14}{3}}}={\sqrt {56}}$ which fortunately is all quite easy to do by hand. So clearly the method suggested by Rckrone is the way you're supposed to do it in this case. Widener (talk) 06:42, 8 February 2012 (UTC)

The no homework policy in my opinion is pretty stupid. What is the different between people who don't know something and people who don't know how to do their homework? This should be the place where everyone gets help regardless of whatever. I don't think people can really exploit this to do homework for them. First of all, homework is not just 1 problem but usually ranging from 10 up to 50 problems. Plus when you asked it here, you're not going to get the answer right the way. You probably have to wait few hours to just get the answer for ONE problem, it would takes forever (i meant they won't have enough time to finish it) for someone to try to do homework base on this. This desk reference can not be exploited as a tool to do someone else's homework anyway even if someone wants to. And I see no problem with helping someone on their homework with 1 or 2 problems. At least give them some direction to head to. So i don't think people here should even bother to ask is this homework because it doesn't do anything good. Plus people can always lie if they want to. The no homework policy here is just useless to be honest.

"You can solve for the lengths of the remaining legs of the right triangles x,y,z by setting up the three equations x+y = 27, x+z = 26, y+z = 25" I don't really understand this part. Can you explain in more details? Which right triangles are you talking about? What segments are you talking about? Name all of them please. And what is x, y, z stand for. By the way, is this base off from angle bisector theorem Pendragon5 (talk) 21:59, 8 February 2012 (UTC)

Hopefully this diagram makes it clearer: [1] Widener (talk) 00:53, 9 February 2012 (UTC)

IT IS! Thanks! Anyway i wonder what theorem is that property come from?Pendragon5 (talk) 01:19, 9 February 2012 (UTC)

What property? Widener (talk) 01:39, 9 February 2012 (UTC)

The property that enable us to set up three equations like this: x+y = 27, x+z = 26, y+z = 25. Pendragon5 (talk) 03:48, 9 February 2012 (UTC)

The reason we know that the two segments labelled as having length x both have the same length is that they are the corresponding sides of congruent triangles. We know those two triangles are congruent because they share the same hypotenuse and they have the same angle at vertex A (because the hypotenuse is the angle bisector). Similarly for the segments labelled y and z. Rckrone (talk) 04:22, 9 February 2012 (UTC)

Oh wow, pretty much common sense.Pendragon5 (talk) 19:20, 9 February 2012 (UTC)

About the homework policy: First is that people are allowed to ask for help with their homework questions. Reference deskers will guide people in the right direction and help them understand the material. What we won't do is hand out answers so they can get around grasping the material. Second, you are wrong about people being able to exploit the reference desk if we did give out answers. You're assuming that everyone's homework is structured like yours, which is not true. For instance in a college or graduate class, problem sets often consist of far fewer but more difficult proof questions. In some ways the question you asked is much more like these problems than the homework you're used to (although the subject matter is not something you'd typically see a class on). Rckrone (talk) 04:34, 9 February 2012 (UTC)

I didn't say people can. I meant they can't exploit it even if they want to because of the waiting time needed to received the answer. And of course beside that i agree with your points. That's also my point of view too anyway.Pendragon5 (talk) 19:18, 9 February 2012 (UTC)