How to measure the 0-100 Km/h in an Auto E60?
09-14-2005, 06:28 AM
#11
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I think the results you are looking for are achievable.
But,
Many of the official results come for professional drivers witht he cars at optimal weight (gas, driver, etc.) and optimal track conditions.
A track is more suitable and traction friendly due to the excessive amount of rubber left behind.
Also, these times reflect the constant beating on these cars.
The mileage and many other factors make a huge difference in their overall performance.
You may see additional reduction in time as the car breaks in.
I would be very interested to see your results.
I would be careful of too much power braking off the line.
This is only to achieve maximum launch with minimum whell spin.
my opinion only.
But,
Many of the official results come for professional drivers witht he cars at optimal weight (gas, driver, etc.) and optimal track conditions.
A track is more suitable and traction friendly due to the excessive amount of rubber left behind.
Also, these times reflect the constant beating on these cars.
The mileage and many other factors make a huge difference in their overall performance.
You may see additional reduction in time as the car breaks in.
I would be very interested to see your results.
I would be careful of too much power braking off the line.
This is only to achieve maximum launch with minimum whell spin.
my opinion only.
09-14-2005, 06:33 AM
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Originally Posted by Shebs' date='Sep 14 2005, 06:07 AM
Hi all,
To have the most powerful start for racing or for measuring 0-100, I do the following:
1. Put the tranny in D, with the DSC on.
I would turn both DSC and DTC off. Be sure you are on a surface that provides good traction. Concrete generally is decent.
2. Press brake pedal with the left foot, while applying throttle with the right foot. My car doesn't seem to be able to rev past 2300 rpm (it is a 520i). Is this normal?
This technique is pretty standard and good for fast starts, but not great on the tranny, etc..
3. Release the brake and get a decent powerful start.
The best 0-100 time I was able to record using this method was around 10.7, although the owner's manual specifies that my car is capable of 9.9.
How are you measuring ET? It is almost impossible to get an accurate measurement using, for example, a stopwatch. And, G meters, while decent, are rather inaccurate. Also, are you allowing for rollout? If not, then your ET will differ from those measured using, say, a timing light on a drag strip.
So I thought of the following techniques:
1. Rev the engine in N, then put the tranny in D. I think this is not good for auto transmission , I never tried this.
I wouldn't.
2. Disengage DSC. I tried this but it seems to offer less acceleration as the power is dissipated through wheel spinning.
Some spinning is optimal generally speaking. Excessive spinning needs to be lessened. Interesting though; I almost never get any wheel spin in my auto 545i. What kind of surface are you on when testing?
3. Just press the accelerator without holding the brake pedal with left foot. Tried this but the car takes sometime to start moving.
That's what I do, but I don't want to hurt the car.
Your comments on the points mentioned above would be greatly appreciated. I really would like to see my car jump to 100 in less than 10 sec.
Perhaps you need to go to a track with a timing light and to get some advice on obtaining the best times. I am confident that BMW's figures are acheivable. Magazine tests actually often yield better results than those published by manufacturers..
Thanks in advance.
To have the most powerful start for racing or for measuring 0-100, I do the following:
1. Put the tranny in D, with the DSC on.
I would turn both DSC and DTC off. Be sure you are on a surface that provides good traction. Concrete generally is decent.
2. Press brake pedal with the left foot, while applying throttle with the right foot. My car doesn't seem to be able to rev past 2300 rpm (it is a 520i). Is this normal?
This technique is pretty standard and good for fast starts, but not great on the tranny, etc..
3. Release the brake and get a decent powerful start.
The best 0-100 time I was able to record using this method was around 10.7, although the owner's manual specifies that my car is capable of 9.9.
How are you measuring ET? It is almost impossible to get an accurate measurement using, for example, a stopwatch. And, G meters, while decent, are rather inaccurate. Also, are you allowing for rollout? If not, then your ET will differ from those measured using, say, a timing light on a drag strip.
So I thought of the following techniques:
1. Rev the engine in N, then put the tranny in D. I think this is not good for auto transmission , I never tried this.
I wouldn't.
2. Disengage DSC. I tried this but it seems to offer less acceleration as the power is dissipated through wheel spinning.
Some spinning is optimal generally speaking. Excessive spinning needs to be lessened. Interesting though; I almost never get any wheel spin in my auto 545i. What kind of surface are you on when testing?
3. Just press the accelerator without holding the brake pedal with left foot. Tried this but the car takes sometime to start moving.
That's what I do, but I don't want to hurt the car.
Your comments on the points mentioned above would be greatly appreciated. I really would like to see my car jump to 100 in less than 10 sec.
Perhaps you need to go to a track with a timing light and to get some advice on obtaining the best times. I am confident that BMW's figures are acheivable. Magazine tests actually often yield better results than those published by manufacturers..
Thanks in advance.
[snapback]170359[/snapback]
09-14-2005, 06:39 AM
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Originally Posted by ls78000' date='Sep 14 2005, 07:50 AM
Also be aware that the 0-100 secs in BMW documents are generally for manual gearbox, not auto. You usually "loose" between 0.5 and 1 sec with an auto gearbox.
[snapback]170384[/snapback]
09-14-2005, 06:46 AM
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1. Line the 520 up three feet from the edge of a large cliff.
2. Tap throttle lightly.
3. When the car feels like its falling effortlessly toward Earth, hit stopwatch.
4. Ouch.
2. Tap throttle lightly.
3. When the car feels like its falling effortlessly toward Earth, hit stopwatch.
4. Ouch.
09-14-2005, 08:56 AM
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Originally Posted by Evenflow545' date='Sep 14 2005, 10:46 PM
1.? ? Line the 520 up three feet from the edge of a large cliff.
2.? ? Tap throttle lightly.?
3.? ? When the car feels like its falling effortlessly toward Earth, hit stopwatch.
4.? ? Ouch.
2.? ? Tap throttle lightly.?
3.? ? When the car feels like its falling effortlessly toward Earth, hit stopwatch.
4.? ? Ouch.
[snapback]170432[/snapback]
LOL give us humble 520, 523 owners a break 
09-14-2005, 10:07 AM
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Originally Posted by Evenflow545' date='Sep 14 2005, 12:53 PM
I know there are some rocket scientists on board.? ?
How fast would a 520i do 0-100 kph at terminal velocity?
How fast would a 520i do 0-100 kph at terminal velocity?
[snapback]170526[/snapback]
heres a good definition, anyone want to do the math
"....it's really an approximate equation for terminal speed (I say speed because velocity describes both speed and direction).
The two dominant forces in free-falling through the atmosphere are drag and
gravity. (Atmospheric) Drag is due to what we call (skin or surface)
friction. Long story short, the typical formula for atmospheric drag is
D = (1/2)*C*A*p*(V^2)
D is the drag force, in Newtons (or pounds)
C is the coefficient of drag, or drag coefficient, which is unitless
C is often written as Cd where "d" is a subscript
A is the area of the body's surface over which the air is flowing
p is the density of the air, in kg/(m^3) or slugs per cubic foot
V is the speed of the air flow relative to the body which is experiencing
the drag
V is expressed in meters per second (m/s) or feet per second
The units for D is then ( kg/(m^3) * (m^2) * (m/s)^2 ) = kg*m/(s^2) =
Newtons
or ( slugs/(ft^3) * (ft^2) * (ft/s)^2 ) = slugs*ft/(s^2) = pounds
note: a slug is the imperial unit for mass, just like the kilogram (kg) is
the SI unit for mass
note: the density of air is not constant, it changes over time and
location. You can read more about this in the references, below.
The force exerted by gravity is called weight. The formula is
W = m*g
W is the weight, in Newtons (or pounds)
m is the mass of the object, in kg (or slugs)
g is the acceleration due to gravity, in m/(s^2) (or ft/(s^2))
note: I'm using the carat "^" for exponentiation. a^b is "a" raised to the
power of "b", so that 2^2 is 4, 4^2 is 16
note: acceleration due to gravity changes with location. This is often
ignored for problems like what you're describing, but if you double your
distance from the center of Earth, then the acceleration due to gravity is
one-fourth what you felt before.
If there were no atmosphere to slow you down, the force of gravity would
continually accelerate you until you collide with something solid (like the
earth). However, the atmosphere does exists and it retards your motion
with the force of drag. Therefore, the resultant force you feel towards
the earth is weight minus drag (W-D) and when your speed increases to the
point that the drag you feel equals your weight, you will accelerate no
more. This then, is your terminal speed. And the equation to find it is:
W = D
m*g = (1/2)*C*A*p*(V^2)
and we can rearrange this with some algebraic manipulation to get
V^2 = 2*m*g/(C*A*p)
or
V = ( 2*m*g/(C*A*p) )^(1/2)"
09-14-2005, 01:05 PM
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Originally Posted by m630' date='Sep 14 2005, 01:07 PM
[quote name='Evenflow545' date='Sep 14 2005, 12:53 PM']I know there are some rocket scientists on board.? ??
How fast would a 520i do 0-100 kph at terminal velocity?
How fast would a 520i do 0-100 kph at terminal velocity?
[snapback]170526[/snapback]
heres a good definition, anyone want to do the math
"....it's really an approximate equation for terminal speed (I say speed because velocity describes both speed and direction).
The two dominant forces in free-falling through the atmosphere are drag and
gravity. (Atmospheric) Drag is due to what we call (skin or surface)
friction. Long story short, the typical formula for atmospheric drag is
D = (1/2)*C*A*p*(V^2)
D is the drag force, in Newtons (or pounds)
C is the coefficient of drag, or drag coefficient, which is unitless
C is often written as Cd where "d" is a subscript
A is the area of the body's surface over which the air is flowing
p is the density of the air, in kg/(m^3) or slugs per cubic foot
V is the speed of the air flow relative to the body which is experiencing
the drag
V is expressed in meters per second (m/s) or feet per second
The units for D is then ( kg/(m^3) * (m^2) * (m/s)^2 ) = kg*m/(s^2) =
Newtons
or ( slugs/(ft^3) * (ft^2) * (ft/s)^2 ) = slugs*ft/(s^2) = pounds
note: a slug is the imperial unit for mass, just like the kilogram (kg) is
the SI unit for mass
note: the density of air is not constant, it changes over time and
location. You can read more about this in the references, below.
The force exerted by gravity is called weight. The formula is
W = m*g
W is the weight, in Newtons (or pounds)
m is the mass of the object, in kg (or slugs)
g is the acceleration due to gravity, in m/(s^2) (or ft/(s^2))
note: I'm using the carat "^" for exponentiation. a^b is "a" raised to the
power of "b", so that 2^2 is 4, 4^2 is 16
note: acceleration due to gravity changes with location. This is often
ignored for problems like what you're describing, but if you double your
distance from the center of Earth, then the acceleration due to gravity is
one-fourth what you felt before.
If there were no atmosphere to slow you down, the force of gravity would
continually accelerate you until you collide with something solid (like the
earth). However, the atmosphere does exists and it retards your motion
with the force of drag. Therefore, the resultant force you feel towards
the earth is weight minus drag (W-D) and when your speed increases to the
point that the drag you feel equals your weight, you will accelerate no
more. This then, is your terminal speed. And the equation to find it is:
W = D
m*g = (1/2)*C*A*p*(V^2)
and we can rearrange this with some algebraic manipulation to get
V^2 = 2*m*g/(C*A*p)
or
V = ( 2*m*g/(C*A*p) )^(1/2)"
[snapback]170528[/snapback]
[/quote]Simple! kind hard to read because i'm not used to read that kind of algebric format...but simple. It's all about some elementary fizics and some math. Right.
09-14-2005, 01:10 PM
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Exactly, M630. I've crunched your figures, and it appears that a 520i will do 0-100kph in exactly 10.7 seconds while plummeting off a cliff.
Shebs, it's safer to keep doing neutral drops. And, if I'm ever cruising around in my rig out in Cairo and stumble along on your transmission, I'll gladly return it.
Shebs, it's safer to keep doing neutral drops. And, if I'm ever cruising around in my rig out in Cairo and stumble along on your transmission, I'll gladly return it.