Threshold speed for stale red light

[California98Civic]

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.

[redpoint5]

Originally Posted by cliffordlj

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.

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?

[vskid3]

According to the Wikipedia article on kinetic energy:

Originally Posted by Wikipedia

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force. As a consequence of this quadrupling, it takes four times the work to double the speed.

[redpoint5]

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.

[JockoT]

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.

[teoman]

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.

[stefanv]

Originally Posted by redpoint5

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.

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.

[redpoint5]

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?

[JockoT]

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.

[stefanv]

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).