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Old 08-27-2017, 01:56 PM
 
17 posts, read 28,949 times
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Mass is constant(minus the gas used).

Momentum increases....=mv
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Old 08-27-2017, 02:41 PM
 
Location: Montgomery County, PA
16,563 posts, read 15,105,033 times
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Quote:
Originally Posted by RP2C View Post
Since mass increases with speed, it takes more energy to accelerate, so you experience that difference. You need to combine that with associated effects at or around 88 mph and frame-of-reference time-dilation.
You forgot the time-space continuum.
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Old 08-27-2017, 05:48 PM
 
887 posts, read 1,204,463 times
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Quote:
Originally Posted by RP2C View Post
Since mass increases with speed, it takes more energy to accelerate, so you experience that difference. You need to combine that with associated effects at or around 88 mph and frame-of-reference time-dilation.
Right. It's that old 'moving clocks go slow" thing so it's really all relative.
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Old 08-27-2017, 05:53 PM
 
33,387 posts, read 34,617,896 times
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Quote:
Originally Posted by Exitus Acta Probat View Post
Is air resistance the only reason a car accelerates at a greater rate from 10-20 mph than it does from 50-60 mph, or is there another reason as well?
air resistance is only one reason. remember that drag forces go up with the square of the speed increase, so drag at 30 miles an hour is something like four times what it is at 20 mile per hour. and it consstantly increase with speed. go 120 mph and you might was well be trying to push a barn through the wind at 35 mph.

add to that rolling resistance, the torque curve falling off as speeds get higher, and horsepower peaking and falling off at higher speeds, and you can see why cars accelerate slower at higher speeds. and as noted, gearing plays a factor here as well.
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Old 08-27-2017, 06:12 PM
 
Location: West Los Angeles and Rancho Palos Verdes
13,563 posts, read 15,485,447 times
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Originally Posted by pvande55 View Post
Simple physics. E =1/2m(v**2). Energy increases by the SQUARE of the speed. It takes four times as much energy to go at twice the speed. If I've lost you since you never had physics in HS, I apologize.
No, you didn't lose me, but it appears my original post lost you. I'm asking what forces are working against the drive wheels at the other side of the vector sum diagram. And my question has been answered.
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Old 08-28-2017, 02:07 AM
 
Location: Eugene, Oregon
11,119 posts, read 5,526,078 times
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Quote:
Originally Posted by Exitus Acta Probat View Post
Is air resistance the only reason a car accelerates at a greater rate from 10-20 mph than it does from 50-60 mph, or is there another reason as well?
At faster speeds, there's greater friction on the tires and internally, all the moving parts of the drive-train have increased friction. But air-resistance is the main factor, as it rises exponentially with speed. If you had an airplane with a shape capable of going twice the speed it can reach with a standard engine, it might take four times the engine thrust to do it (at least).
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Old 08-28-2017, 04:52 AM
 
Location: The Driftless Area, WI
7,112 posts, read 4,947,887 times
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Are you guys ready for the right answer?

In regards the pure physics: F = ma & E = 1/2 mv^2; mix and match and you get a = 2E/ Fv^2--> that means at any given E & F (ie rpm & fuel consumption) then--> a goes down as v goes up.

But that's theory. In practice:

It's gearing. Low gear puts all the power into shorter steps and you can take short steps faster than you take long steps. It's the reason you down shift to come out of a turn and why you qtr-milers don't start out in 4th gear.
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Old 08-28-2017, 08:28 AM
 
3,754 posts, read 4,164,087 times
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Quote:
Originally Posted by Malloric View Post
F=(1/2)*p(v^2)*Cd*A.

Since p, Cd, and A are all constants you're left with v^2.

I guess if 400 and 3,600 is a negligible difference to you, then it's a negligible difference.

That's really a stupid thing to say though. It's a bigger difference than 60 versus 120. Why don't we get fabulous mileage at 20 then? Because the drag at 20 mph in a car is a negligible. By 60 mph that's no longer true. That 900% increase in aerodynamic drag from 20 to 60 has become important in terms of fuel economy. That's why you'll get better gas mileage going 50 than 60 or 60 instead of 70.
LOL. It's a negligible difference at 20, and it's still negligible at 60. Do me a favor... tell me what your numbers mean, hmm? 400 WHAT? 3,600 WHAT? You're wrong about your gas mileage claim as well. 10mph increase in speed can be entirely negated by gearing as far as mpg goes.

Quote:
Originally Posted by Regajohn View Post
Maybe on the Moon it is negligible, but here on Earth where most of us drive it is a huge factor.
Someone else who doesn't know what they're talking about. Fantastic.

Quote:
Originally Posted by guidoLaMoto View Post
Are you guys ready for the right answer?

In regards the pure physics: F = ma & E = 1/2 mv^2; mix and match and you get a = 2E/ Fv^2--> that means at any given E & F (ie rpm & fuel consumption) then--> a goes down as v goes up.

But that's theory. In practice:

It's gearing. Low gear puts all the power into shorter steps and you can take short steps faster than you take long steps. It's the reason you down shift to come out of a turn and why you qtr-milers don't start out in 4th gear.
You understand. But without numbers, I doubt the rest of these posters get it.

The amount of horsepower it takes for an average car with an average frontal area and .35 Cd is about FIVE horsepower to hit 30 mph. Yes, that's right. Five. At 60 mph? You need 20 horsepower.

That's why it's negligible. If you know anything about modding cars, it's that a 15 horsepower difference is a drop in the bucket. You can't feel it. It's too small to matter. The fact of the matter is, at 60mph, the average car, with a Cd of .35, will produce 160 lbs of drag. That amount of resistance is nothing for a car.
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Old 08-28-2017, 12:57 PM
 
Location: Vallejo
21,599 posts, read 24,739,140 times
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Quote:
Originally Posted by Katana49 View Post
LOL. It's a negligible difference at 20, and it's still negligible at 60. Do me a favor... tell me what your numbers mean, hmm? 400 WHAT? 3,600 WHAT? You're wrong about your gas mileage claim as well. 10mph increase in speed can be entirely negated by gearing as far as mpg goes.
Only if you consider a 90-fold increase to be negligible. It's a unit-less measure. Everything else is constant and divides out leaving the square of velocity... which again divides out. As long as you're using the same units (standard or metric) and not comparing km/h and metric to mph and standard it really doesn't matter.

Quote:
Someone else who doesn't know what they're talking about. Fantastic.
No, pretty much everyone gets what they're talking about. The moon has essentially no atmosphere so no air resistance. Mod cut.

Quote:
The amount of horsepower it takes for an average car with an average frontal area and .35 Cd is about FIVE horsepower to hit 30 mph. Yes, that's right. Five. At 60 mph? You need 20 horsepower.
Hence why it's not negligible. Even if we take your numbers, that's clearly true.
1 HP is equivalent to around 2,500 BTU/hours. That is to generate 1 HP for an hour is the equivalent work of 2,500 BTUs of energy. A gallon of gasoline contains 114,000 BTUs.

A gallon of gasoline contains enough energy to produce 5 HP for 9.12 hours.
A gallon of gasoline contains enough energy to produce 20 HP for 2.28 hours.

9.12 hours X 20 mph is 182 miles.
2.28 hours X 60 mph is 136.8 miles.

That's a non-negligible difference of around 33%. Using your own numbers, it's not negligible.

Of course, you're not using appropriate numbers. The 5 and 20 hp figures, which are reasonable for an average car, is for everything. Drag, parasitic loss, rolling resistance. It's not just for aerodynamic drag. As to why you don't get 182 or 136.8 mpg in a car... again, obvious. ICEs are not 100% efficient or even close to. More like 15-20%.

As you can see, it's not negligible. But carry on being wrong. Air resistance is relatively unimportant at low speeds because of simple physics. Rolling resistance is generally linear with speed. Generally, it's not precise. The formula for coefficient of friction is dependent on speed but it's a generally linear relationship. That is it's roughly three times as high at 60 mph than it is at 20 mph. Aerodynamic drag, on the other hand is not roughly three times as high at 60 mph as it is at 20 mph. It's 90 times as high.

By low we're talking very low. By 20 mph, aerodynamic drag has already caught up with rolling resistance. There's tons of other variables. At very low speed, most engines are not operating efficiently (especially larger ones). There's a lot of pumping loss at nearly closed throttle. Of course for a larger engine 5 hp or 20 hp isn't really relevant. A Corvette will loaf along the freeway at 60 mph at maybe 1,350 rpm. I don't know what a Corvette makes at 1,350 rpm. Call it 200 hp, probably more. It's running at or less than 10% throttle. Part of what cylinder deactivation does is bump that to 20% throttle and shut off half the cylinders. Not only does it reduce parasitic losses but there's less pumping loss at 20% throttle than 10%.

It's all the other things which amount to why most cars get optimal fuel economy around 40 mph. In something like my car, I can roll along for quite a while at 20 mph without even using the ICE at all and it's extremely efficient. If I'm going 35 mph and anticipate a lot of low speed driving, I'll play games with the throttle to try and keep the ICE on at 35 rather than deplete the battery as 35 mph will deplete it much more quickly due to most of the drag being from aerodynamics rather than rolling resistance at that speed.

Last edited by PJSaturn; 08-28-2017 at 09:38 PM.. Reason: Personal attack.
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Old 08-28-2017, 01:01 PM
 
9,332 posts, read 6,866,407 times
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Not only at higher speeds but look at the torque and hp power bands within RPMs... You'll notice alot of vehicles peak early or mid in the rpm band and then fall off a cliff. Other vehicles make very little power early in the band and need to be rev'd to high rps (e.g. porsche boxter/911s).
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