Threshold speed for stale red light
 

Every time I cruise up to a stale red light, I try to brake a bit early, allowing me to drive slowly, so that when it turns green, I'm still driving 5-20 MPH. If I stay off the brakes altogether, I find often that I have to brake hard at the last minute, which then brings my speed to (near) zero.

Is there a chart for how much energy you save, by not having to start your car from a full stop? I can only assume that the faster you are going, the less energy it would take. But is there a point that I definitely want to avoid--say, 5 MPH, or 15 MPH--at which point, there is little benefit to cruising at that speed, over starting from a full stop? If I knew what speed was my minimum, I would be better able to time my braking (or not brake).

Sorry if this isn't clear. I can explain in more detail if I need to.

Hey Prr, interesting question. Basically the answer is you want to maintain as much speed as possible. Every 1 mph you slow down is 1 mph you have to spend fuel to speed back up. There is no minimum threshold you want to avoid other than stopping completely. The higher speeds are more 'fuel costly' to regain only due to additional aerodynamic drag. Ignoring aerodynamics, 1 mph is 1 mph. Accelerating from 10-20 mph requires virtually the same amount of energy as accelerating from 40-50 mph. Some frictional losses are increased at higher speeds, but its negligible.

OK, well that answers my question. I will just keep guesstimating, based on how stale I think the light is, how fast I'm going, and how much space there is for me to brake.

OK well at least I don't have to wonder now. Thanks.

What Daxo said.

as far as I'm concerned, it's a pulse & glide situation, with the hopes of avoiding having to use the brakes. (Unless you have regen braking...then you only lose out on whatever inefficiencies there are in the system. )

On my commute, at the lights that are likely to cause me to stop, I count to see how long it takes for the light to cycle. I use this to help judge where I need to let off the gas/how fast I should be going to not have to stop. You should always be trying to maintain as much momentum as you safely/legally can.

Don't be surprised if some people are annoyed by the fact that you are coasting up to a red rather than trying to get there first.

If no one is behind me, I'll start my coastdown really early when I see a yellow or red ahead.

If I misjudge it and the light turns green before I reach it, and so have to accelerate a little to get back up to speed, so be it. Better than getting going from a dead stop!

Wait until you guys get factory stop/ start cars.

Then you have a choice between braking early and maintaining momentum, or braking late and having the engine switch off sooner.

:eek:

As far as what daox said, 1 mph at 5 is not the same as 1 at 40 mph. The kenetic energy scales with the square of the speed. You can illustrate this by coasting up a hill and observing the vertical displacement will be significantly more as you slow from 40 to 20 as it will be from 20 to 0. However if driving a manual anyways keeping the speed above your first gear idle rpm will reduce clutch wear which is good

The juggling act is accurately predicting when the light will change and the cars in front will begin moving. Brake too much, and the light will change before you get there, and you will have to accelerate, and possibly cause people behind you to miss the green light. Brake too late (or too soft), and you might end up stopping at the light.

If you have a very good idea of when the light will change, it's most efficient to brake very hard early on, reducing your speed to what will perfectly coast you to the light as it changes.

Be mindful of those behind you though. Most people are in a race to the red light, and won't be prepared for "some whacko nutjob" who brakes early and coasts to the red light.

Most people seem to see a red light as a "checkered flag".:rolleyes:

In a situation like this, with a stale red light in the distance (and not blocking anyone behind me) I do one of several things. First, how well do I know the light or how well can I tell how stale it is? If i know it well or can tell how stale it might be, I might keep the engine off and glide. My next most common tactic is light regen braking if I have good speed and longish distance to the light: kiss-start with clutch, select a gear to give me RPMs well over 2000, and switch on my alternator while slowing in gear under DFCO. I brake with friction brakes only when I must.

If it takes an increasing amount of energy to gain each increment of speed, then why does gravity have a constant acceleration rate (9.8m/s/s) despite the increasing speed and constant force?

According to the Wikipedia article on kinetic energy:

Still doesn't explain the constant rate of acceleration due to gravity. Gravity can't input more force, and yet you fall at the same rate of acceleration.

Speed of a motor vehicle is a factor of Drag. As you double the Speed, you quadruple the Drag, hence you need 4 times the power. At least that's my understanding of it.

I will try to give you a ballpark figure. I will also try to give you the formulas so if your car weights 3 metric tonnes and you like to drive at 90 mph between traffic lights, you can redo the calculations and come up with your number.


Assuming your are on flat ground
Your car has kinetic energy whilst in motion for which the formula is:
1/2 m*v^2

( v^2 is v to the power of 2, or v squared, or v*v).

where m is mass in kg, v is velocity in m/s. The result is in Joules.

When you stop, you transform all those joules to thermal energy. (Btw 1 Joule = 1Ws) {If you ever wanted to calculate how much you could re-charge your battery from that stop}.


1 kg of gasoline has 46 MJ of energy.
https://en.wikipedia.org/wiki/Energy_density

There are a lot of other variables at play here but these basic formulas will give you a ballpark number.

Lets crunch some numbers.

You state that you are traveling at 40 mph. -> approx 18 m/s.
Lets assume your car weights 1500 kg's.

So your car has (1500 * 18^2) /2 joules of energy. -> 243000J -> 243 kJ
Here is how you can do the calculations on google.
https://www.google.com.tr/search?q=40+mph+in+m%2Fs&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=D8VkWe29OoLSXsLLivAG#q=(1500+*+18^2 )/2


If we assume that your engine is 1/3 efficient in converting chemical energy to mechanical energy. 1 kg of galsoline has 46/3 MJ. -> 15.3 MJ -> 15300kJ

243 / 15300 = 0.016 kg of gasoline. -> 16 grams.

Density of gasoline is 0.726g/cm3

16/0.726 -> 22 ml. (1 ml = 1cm3)

So in conclusion, if we assume that our simplistic model is semi accurate (with this level of calculation, once you are at 40mph you will continue at mph forever because we did not factor in friction and air drag) and we also assume that our car of 1500kg (3300lbs) traveling at 40mph coming to a full stop and then re-accelerating to 40 mph will waste 22 mililiters of gasoline. (between 11 ml and 44 ml {assuming %100 calculations error in both directions})

11 ml -> 0.003 US liquid gallon if we over estaimated by %100
22 ml -> 0.006 US liquid gallon
44 ml -> 0.012 US liquid gallon if we under estimated by %100

You can directly scale these results for your car's weight (i.e. if your car is 3000kg, just double everything). If you want to calculate for 20 mph (which is half of 40 mph) you have to take 1/4 of the value (20/40 = 1/2, (1/2)^2=1/4)


If it is 2000kg and you want to know for 30mph, you have to redo the calculations. But it is not too hard to do.


If you want to calculate for hills etc, the formula is m*g*h where m is mass in kg, g is gravity constant (approx 9.8) and h is height in meters, result also in Joules.

Force isn't the same as energy.

1. Force is mass times acceleration.

For a constant force and a fixed mass, acceleration will be constant. In the case of gravity, the force is proportional to the mass, so acceleration will always be 9.8 m/s/s regardless of the mass.

2. Energy is force times distance.

If you drop an object, it will accelerate at 9.8 m/s/s, and after one second, it will have traveled a total of 4.9m (since it's average speed over that second will have been 4.9 m/s). After two seconds, it will have fallen 19.6m (two seconds times an average speed of 9.8 m/s). After three seconds, it will have fallen 44.1 meters (three seconds times 14.7 m/s), and so on.

Notice that in the above, speed is proportional to time (after N seconds, it is moving at N x 9.8 m/s), and distance is proportional to time squared (after N seconds, it has fallen N x N x 4.9m) Alternatively, you could say that the distance is proportional to speed squared, since speed is proportional to time.

Since energy is force times distance, the force is constant, and distance is proportional to speed squared, we can see that energy is proportional to speed squared.

That all makes sense, but the confusing part is that the engine uses energy to exert a force that changes the velocity.

If the engine consumes a constant amount of energy and exerts a constant force, it produces a linear acceleration.

In other words, Daox was not wrong in stating that it takes the same amount of energy to accelerate from 10-20 MPH, as it does to go from 40-50 MPH.

How does a constant chemical energy conversion into kinetic energy result in exponentially growing kinetic energy?

These theories are all very well IN A VACUUM but once you introduce air you introduce DRAG. The whole reason you streamline your vehicles. In a vacuum a brick has the same drag as an Indy car.

Drag (physics) For a solid object moving through a fluid or gas, drag is the sum of all the aerodynamic or hydrodynamic forces in the direction of the external fluid flow. It therefore acts to oppose the motion of the object, and in a powered vehicle it is overcome by thrust.

Yes, that is correct. And the original question about gravity is also affected that way. You won't keep accelerating if you're falling. For example, a skydiver, before she opens her parachute, will start by accelerating downward at 9.8 m/s/s, but then this acceleration starts to slow, typically maxing out at about 200 km/h. At that point the upward force of drag (based on her speed, surface area, and drag coefficient) will equal the force of gravity (based on her mass, and the Earth's gravity, which is approximately constant).

You would be losing more energy to drag and friction if you never stopped / slowed down for the stoplight.

For the amount of gasoline lost, i (personally) would still stick with what i found. That is a good enough approximation in my opinion.

Just to ponder on the question a bit further...

The unknowns were friction and air drag, these increase with speed, as you are below the set speed these will always be lower than stopped or accelerating.

Lets still assume that the guy was traveling at 40 mph.
And lets assume that he consumes 4l/100km @ 40 mph.

All of the energy in the fuel is going to combat air drag and other frictions. As the other dude mentioned if we were in space, once you accelerated to 40mph you would continue untill infinity.


Lets assume that you stopped in 200 m and re-accelerated in 300m. So total distance is 500m.

4000ml / (100.000) = 0.04 ml of fuel per meter.

We pulled 500m from a well calibrated deep dark datasource...
500m * 0.04ml/m = 20ml of fuel consumed in 500m.

If we assume that the speed was linearly decreased and linearly increased and drag and friction are linear with speed (which they are absolutely NOT, but for a slow speed of 40mph we can assume that it is without introducing an error of several orders of magnitude).

Then the total drag and friction is exactly half of that value... so 10ml of fuel.

Obviously you meed to plug in the appropriate numbers. So you wasted a total of 11ml + 10ml of fuel for the stop and go.

Had you not slowed down at all, you would have used only 20ml of fuel.

Bu you did travel at 20mph for 500m of the route and lost time.

Obviously you need proper data sources for these calculations.


(And had you traveled at 20mph all along, you would have consumed 10ml of fuel in total, so you wasted 11 ml from a general efficiency perspective, apples to apples comparison where total distance and time are equal).

Drag has nothing to do with the paradox of expending a linear amount of chemical energy, and gaining an exponential amount of kinetic energy.

Something isn't adding up correctly here. I maintain that Daox made no error in his reply.

That is not what i said.

Conservation of energy, it cannot dissapear, so all the fuel you are using while at steady speed is effectively going to drag and friction.

I'll post my question in a physics forum.

The energy to accelerate from zero to a speed (or decelerate to zero from a speed) is 1/2 X mass X speed^2. We start by ignoring friction, hills, driveline losses, engine inefficiency, and air drag. An ideal engine that accelerates a vehicle to a speed will burn an amount of fuel proportional to the kinetic energy.

Sample calculation:
Assume a 3000 lb vehicle.
At zero speed, the kinetic energy is zero.

At 10 MPH, the kinetic energy is 0.5 X 3000 lbs / 32.2 ft/sec^2 X (10 mi/hr X 5280 ft/mi / 3600 sec/hr)^2 = 10,020 ft-lbs kinetic energy.

At 20 MPH, the kinetic energy is 0.5 X 3000 / 32.2 X (20 X 5280 / 3600)^2 = 40,080 ft-lbs.

To accelerate from zero to 10 MPH takes 10,020 ft-lbs of energy.
To accelerate from zero to 20 MPH takes 40,080 ft-lbs of energy.
To accelerate from 10 MPH to 20 MPH takes 40,080 - 10,020 = 30,060 ft-lbs of energy.
Accelerating from 10 to 20 takes three times as much energy as accelerating from 0 to 10.

I'd let that go as a talking point.

stefanv does a good job with the Newtonian physics, but consider: https://en.wikipedia.org/wiki/Potential_energy

Gravity is poorly understood, it is a side effect of the electric nature of the universe, like the Casimir effect.

Back ontopic, I like shooting past cars immobilized at the light as much as anyone, but just after I got my Superbeetle, I slowed for a red light and a black Jetta behind me sped up, raced past me, pulled back into my lane, realized the light was red, jammed the brakes, went hammer down when it turned green, and then realized I was right on his tail and he brake-checked me. Like I was invading his personal space.

I hit him square, so the hood was damaged but not the headlights. He pulled over but when I asked about his insurance he remembered he had somewhere to be.

Grrr

I have no problem understanding the math, but rather the paradox that reality shows us.

A rocket in space with a constant weight and thrust will accelerate at a linear rate. Somehow it gains kinetic energy exponentially, but expends energy linearly.

From experience, I can tell you it doesn't take 3x as long to accelerate from 10-20 MPH as it does from 0-10 MPH. This suggests it doesn't take 3x the energy.

Perhaps there is a correlation between people who don't plan ahead when approaching red lights, and people who don't plan ahead to get insurance.

It's too bad you didn't have a cattle guard on that Beetle. Maybe he would have learned the lesson then.

If you accelerate a land vehicle at constant acceleration, it will take the same time to accelerate from 0 to 10 as from 10 to 20. But the acceleration from 10 to 20 will take three times the distance as the acceleration from 0 to 10. Same force times triple the distance equals three times the energy.

A wheel driven land vehicle generates force by pushing against a (relatively) infinite mass, the Earth. A rocket is different. It generates force by throwing mass in the opposite direction at high velocity. A properly geared wheel driven land vehicle can generate large force at near zero speed using very little power. A rocket uses huge power to generate force at low speed, so has extremely low efficiency.

Reality doesn't show us paradoxes. Those are all in [y]our head[s].

Wheels vs rockets is beside the point. In a gravitational gradient you're trading potential for kinetic energy.