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One factor in your better gas milage out west might be the lack of the additives in the gasoline that are required by EPA in certain areas of the country to reduce pollution.

Buz Groshong wrote:One factor in your better gas milage out west might be the lack of the additives in the gasoline that are required by EPA in certain areas of the country to reduce pollution.

Primarily two types of additives required by the EPA, those that act as detergents and are required throughout the country (these help with mileage long term but should not have a short term impact) and those that reduce pollution by adding oxygen to gasoline (the additive used for this is currently ethanol).

Day Hiker wrote:On a grade, the power-to-fuel efficiency would actually be higher, assuming speed and gear ratio (thus engine speed) remains constant. This is because of the relationship between fuel and power (roughly y=mx+b) that I mentioned before, where y is the fuel usage and x is the engine's load, at a given rpm. It's not proportional (b>0), but it is roughly linear (x^1) for a typical engine. So at a given engine speed, at higher power output, the engine is more efficient, basically because it takes a certain amount of power to just spin the engine, under no load. So the higher the load, the higher the proportion of fuel energy is going to the load.

Yes makes sense. And we could break down the power requirements into maybe 3 portions? Internal friction / resistance (F), air drag (D), and gravity (G). Total power needed to generate T = F + D + G. Here's the comparison: One car drives 40 miles at 80 mph flat. Another drives 20 miles up 5% grade at 80 and down 20 miles at 5% at 80mph.

In both cases, the same amount of internal friction and and air drag forces must be overcome (F+D). In the uphill / downhill case, there is also work against gravity (i.e external work) uphill. Downhill, gravity assists against friction & drag. (+G, then -G).

At 80 mph downhill, as long as gravitational force is less than friction and drag forces G < (F+D), then energy won't be lost through braking. The +/- gravitational work will cancel out, and you should have ~ the same energy required in both cases.

However, as others are saying, drag will vary with altitude, so if all else is equal then this would reduce drag in the uphill / downhill case.

Ze wrote:

Day Hiker wrote:On a grade, the power-to-fuel efficiency would actually be higher, assuming speed and gear ratio (thus engine speed) remains constant. This is because of the relationship between fuel and power (roughly y=mx+b) that I mentioned before, where y is the fuel usage and x is the engine's load, at a given rpm. It's not proportional (b>0), but it is roughly linear (x^1) for a typical engine. So at a given engine speed, at higher power output, the engine is more efficient, basically because it takes a certain amount of power to just spin the engine, under no load. So the higher the load, the higher the proportion of fuel energy is going to the load.

Yes makes sense. And we could break down the power requirements into maybe 3 portions? Internal friction / resistance (F), air drag (D), and gravity (G). Total power needed to generate T = F + D + G. Here's the comparison: One car drives 40 miles at 80 mph flat. Another drives 20 miles up 5% grade at 80 and down 20 miles at 5% at 80mph.

In both cases, the same amount of internal friction and and air drag forces must be overcome (F+D). In the uphill / downhill case, there is also work against gravity (i.e external work) uphill. Downhill, gravity assists against friction & drag. (+G, then -G).

At 80 mph downhill, as long as gravitational force is less than friction and drag forces G < (F+D), then energy won't be lost through braking. The +/- gravitational work will cancel out, and you should have ~ the same energy required in both cases.

However, as others are saying, drag will vary with altitude, so if all else is equal then this would reduce drag in the uphill / downhill case.

Everything there makes sense.

Here is another thought: The total work involving the variable F is equal for both situations, assuming the engine speed is the same for both situations. So for the comparison above, the total work performed on this task is equal (if the speeds are the same and the total gear ratio is the same). But if the downhill is such that you can spend some time coasting in neutral, the engine speed goes to idle for these periods.

So the hill case actually has potential benefit over the flat case, provided you don't downshift on the uphill and no braking is required on the downhill, as you mentioned. With a manual transmission, the downshifting won't occur, unless of course the uphill is so steep it is unavoidable. With a failomatic (Ameritard) transmission, all bets are off.

I put my car in neutral all the time on downhills, if 6th gear is providing too much engine braking. There is even a nice hill on I-515 here in Henderson that allows you to coast in neutral at just the right freeway speed.

But like I mentioned earlier, that's illegal here in Nevada, though.