How much difference does aerodynamics make?
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From: Jtown, Il
The easiest way to get an idea of the drag FORCE is to take your car up to speed, for example 110 mph, let it decel maybe 5mph. Time your 5 mph decel and now you can find an average force.
D=m*("delta" V)/("delta t)
D=Drag force
m=car mass
"delta" V=change in velocity
"delta" t= Time to decel
Taking what mikegarrison stated we can calc. a rough drag coefficient
Cd=D/(1/2*rho*V^2*A)
Cd= Drag coefficient
The trick is to measure the frontal area of the car (the profile, not the actual surface area)
This is rough, but the decel experiment will at least let you compare cars for a given speed range.
D=m*("delta" V)/("delta t)
D=Drag force
m=car mass
"delta" V=change in velocity
"delta" t= Time to decel
Taking what mikegarrison stated we can calc. a rough drag coefficient
Cd=D/(1/2*rho*V^2*A)
Cd= Drag coefficient
The trick is to measure the frontal area of the car (the profile, not the actual surface area)
This is rough, but the decel experiment will at least let you compare cars for a given speed range.
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From: San Fran
Originally Posted by i_heart_my_DB8,Feb 3 2006, 12:38 PM
On the other hand, negative lift is the action of air pulling the car down. Think of the flat bottom of a Ferrari 360 Modena, and the upward pointed exit channels at the rear of the car. The underside of the car is shaped like an upside-down wing. As air rushes underneath the car, it creates a low-pressure zone between the car and the ground (think bernoulli's principle), and that pulls or sucks the car down. This instance is referred to as negative lift.
Or correct me if I'm wrong
.
Or correct me if I'm wrong
.Less air travels beneath the car, than above the car. With a flat underbody and diffusers , air is allowed to flow through relatively undisturbed and then as it reaches the diffuser it speeds up. This creates the low pressure zone under the car.
Correct me if I'm wrong mike.
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From: Scatterbrainia
Originally Posted by NEODYMIUM,Feb 3 2006, 11:45 AM
The easiest way to get an idea of the drag FORCE is to take your car up to speed, for example 110 mph, let it decel maybe 5mph. Time your 5 mph decel and now you can find an average force.
D=m*("delta" V)/("delta t)
D=Drag force
m=car mass
"delta" V=change in velocity
"delta" t= Time to decel
Taking what mikegarrison stated we can calc. a rough drag coefficient
Cd=D/(1/2*rho*V^2*A)
Cd= Drag coefficient
The trick is to measure the frontal area of the car (the profile, not the actual surface area)
This is rough, but the decel experiment will at least let you compare cars for a given speed range.
D=m*("delta" V)/("delta t)
D=Drag force
m=car mass
"delta" V=change in velocity
"delta" t= Time to decel
Taking what mikegarrison stated we can calc. a rough drag coefficient
Cd=D/(1/2*rho*V^2*A)
Cd= Drag coefficient
The trick is to measure the frontal area of the car (the profile, not the actual surface area)
This is rough, but the decel experiment will at least let you compare cars for a given speed range.
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From: St-Redempteur,Qc.
Originally Posted by mikegarrison,Feb 3 2006, 01:42 PM
The answer to this is a little more complicated than you might think.
Drag is proportional to the size of the object and the "dynamic pressure" (1/2 * air_density * velocity^2). This usually ends up being descibed as:
D = 1/2 * rho * V^2 * S * Cd
Where rho is air density, V is velocity, S is some reference area, and Cd is a non-dimensional number called the Coefficient of Drag. For cars, S is usually taken to be the "frontal area".
So the Cd is a measurement of how efficient the shape is, independently of the size of the vehicle. That means if we have the same Cd as an SUV, we can still have less drag because we are smaller. But if a coupe of the same size has a better Cd, then it will have less drag.
Lift (or negative lift, aka downforce) has a similar equation, except it has a Cl instead of Cd.
The ratio of Cl/Cd is one of the key measures for an airplane of "how aerodynamic" it is. For an airplane, you want Cl to be high and Cd to be low. But in the case of a street car you probably want Cl to be close to zero and Cd to just be as small as you can get it.
ps. Cd (and Cl) are not constants! They depend on another factor called Reynolds Number, which changes with speed. There are also different kinds of drag, including drag which is caused by lift. So it's only an approximation to say that the drag is a fixed number that is proportional to V^2, but it's a pretty good approximation. And them you have to factor in mechanical friction etc., so the whole thing gets complicated quickly.
Drag is proportional to the size of the object and the "dynamic pressure" (1/2 * air_density * velocity^2). This usually ends up being descibed as:
D = 1/2 * rho * V^2 * S * Cd
Where rho is air density, V is velocity, S is some reference area, and Cd is a non-dimensional number called the Coefficient of Drag. For cars, S is usually taken to be the "frontal area".
So the Cd is a measurement of how efficient the shape is, independently of the size of the vehicle. That means if we have the same Cd as an SUV, we can still have less drag because we are smaller. But if a coupe of the same size has a better Cd, then it will have less drag.
Lift (or negative lift, aka downforce) has a similar equation, except it has a Cl instead of Cd.
The ratio of Cl/Cd is one of the key measures for an airplane of "how aerodynamic" it is. For an airplane, you want Cl to be high and Cd to be low. But in the case of a street car you probably want Cl to be close to zero and Cd to just be as small as you can get it.
ps. Cd (and Cl) are not constants! They depend on another factor called Reynolds Number, which changes with speed. There are also different kinds of drag, including drag which is caused by lift. So it's only an approximation to say that the drag is a fixed number that is proportional to V^2, but it's a pretty good approximation. And them you have to factor in mechanical friction etc., so the whole thing gets complicated quickly.
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From: Covington WA, USA
Originally Posted by i_heart_my_DB8,Feb 3 2006, 11:38 AM
Or correct me if I'm wrong
In fact, there is no such thing as "negative pressure" so the air ALWAYS pushes on the car. The only question is whether it pushes more on the bottom than the top (positive lift) or whether it pushes more on the top than the bottom (negative lift). Downforce is just another name for negative lift -- assuming you are talking about aerodynamics. If you are talking about the vertical total force on the car, that would be the force due to gravity in addition to the force due to lift.
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From: Covington WA, USA
Originally Posted by pantyraider,Feb 3 2006, 11:49 AM
I've always thought of flat underbodies and diffusers as the under side of a huge wing, where the wing is the car itself.
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From: Scatterbrainia
Originally Posted by mikegarrison,Feb 3 2006, 12:23 PM
You are wrong. It's all the same.
In fact, there is no such thing as "negative pressure" so the air ALWAYS pushes on the car. The only question is whether it pushes more on the bottom than the top (positive lift) or whether it pushes more on the top than the bottom (negative lift). Downforce is just another name for negative lift -- assuming you are talking about aerodynamics. If you are talking about the vertical total force on the car, that would be the force due to gravity in addition to the force due to lift.
In fact, there is no such thing as "negative pressure" so the air ALWAYS pushes on the car. The only question is whether it pushes more on the bottom than the top (positive lift) or whether it pushes more on the top than the bottom (negative lift). Downforce is just another name for negative lift -- assuming you are talking about aerodynamics. If you are talking about the vertical total force on the car, that would be the force due to gravity in addition to the force due to lift.
But in the examples I used: The Sprint car with the huge wing vs. the 360 Modena with the sculpted/flat underbody, is the same aerodynamic principle being utilized? I realize the end result is the same: at speed, differences in air pressure cause the car to be drawn to the center of the earth. But is there a differentiation between whether this is caused by increased air pressure from above vs. decreased air pressure from below?
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From: Covington WA, USA
Originally Posted by Emil St-Hilaire,Feb 3 2006, 12:14 PM
Mike,I'm sure,you're an Engineer.!!!
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From: Milwaukee Area
Originally Posted by mikegarrison,Feb 3 2006, 12:42 PM
The answer to this is a little more complicated than you might think.
Drag is proportional to the size of the object and the "dynamic pressure" (1/2 * air_density * velocity^2). This usually ends up being descibed as:
D = 1/2 * rho * V^2 * S * Cd
Where rho is air density, V is velocity, S is some reference area, and Cd is a non-dimensional number called the Coefficient of Drag. For cars, S is usually taken to be the "frontal area".
So the Cd is a measurement of how efficient the shape is, independently of the size of the vehicle. That means if we have the same Cd as an SUV, we can still have less drag because we are smaller. But if a coupe of the same size has a better Cd, then it will have less drag.
Lift (or negative lift, aka downforce) has a similar equation, except it has a Cl instead of Cd.
The ratio of Cl/Cd is one of the key measures for an airplane of "how aerodynamic" it is. For an airplane, you want Cl to be high and Cd to be low. But in the case of a street car you probably want Cl to be close to zero and Cd to just be as small as you can get it.
ps. Cd (and Cl) are not constants! They depend on another factor called Reynolds Number, which changes with speed. There are also different kinds of drag, including drag which is caused by lift. So it's only an approximation to say that the drag is a fixed number that is proportional to V^2, but it's a pretty good approximation. And them you have to factor in mechanical friction etc., so the whole thing gets complicated quickly.
Drag is proportional to the size of the object and the "dynamic pressure" (1/2 * air_density * velocity^2). This usually ends up being descibed as:
D = 1/2 * rho * V^2 * S * Cd
Where rho is air density, V is velocity, S is some reference area, and Cd is a non-dimensional number called the Coefficient of Drag. For cars, S is usually taken to be the "frontal area".
So the Cd is a measurement of how efficient the shape is, independently of the size of the vehicle. That means if we have the same Cd as an SUV, we can still have less drag because we are smaller. But if a coupe of the same size has a better Cd, then it will have less drag.
Lift (or negative lift, aka downforce) has a similar equation, except it has a Cl instead of Cd.
The ratio of Cl/Cd is one of the key measures for an airplane of "how aerodynamic" it is. For an airplane, you want Cl to be high and Cd to be low. But in the case of a street car you probably want Cl to be close to zero and Cd to just be as small as you can get it.
ps. Cd (and Cl) are not constants! They depend on another factor called Reynolds Number, which changes with speed. There are also different kinds of drag, including drag which is caused by lift. So it's only an approximation to say that the drag is a fixed number that is proportional to V^2, but it's a pretty good approximation. And them you have to factor in mechanical friction etc., so the whole thing gets complicated quickly.
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From: Milwaukee Area
Originally Posted by mikegarrison,Feb 3 2006, 02:29 PM
Aero engineer, yes. So I know more about airplane aerodynamics than cars. But it's a case where the theory is all the same, but some of the applications of the theory are different.
