Quote, originally posted by AlecGTI » 
To truly limit brake dive beyond just modulating the rear brakes to drag the rear a bit, the car would need active suspension. Last time I checked, we don't have active air or hydro springs or dampers :( 
Quote, originally posted by BigBlockBug » 
no its not search for the active susension from Bose. Yes Bose, do a search for "learning center" and Bose suspension. thats all I remember there are some vids that prove it is possible

Quote, originally posted by BigBlockBug » 
download the video then tell me that, im not saying it eliminates it completely, there is almost never such a thing as 100% in physics, but its damn close from what i can see 
Quote, originally posted by AlecGTI » 
If I may weigh in again: While a braking system cannot really help brake dive and other undesirables, a suspension can. And they didn't put weights in the car to make the suspension look better, that accusation is pure ignorance. 
Quote, originally posted by Website from above... » 
Newtons laws apply here. Really the third one is all we need to worry about right now. The third law: Every force on a car by another object, such as the ground, is matched by an equal and opposite force on the object by the car. When you apply the brakes, you cause the tires to push forward against the ground, and the ground pushes back. As long as the tires stay on the car, the ground pushing on them slows the car down. Let us continue analyzing braking. Weight transfer during accelerating and cornering are mere variations on the theme. We won't consider subtleties such as suspension and tire deflection yet. These effects are very important, but secondary. The figure shows a car and the forces on it during a ``one g'' braking maneuver. One g means that the total braking force equals the weight of the car, say, in pounds. In this figure, the black and white ``pie plate'' in the center is the CG. is the force of gravity that pulls the car toward the center of the Earth. This is the weight of the car; weight is just another word for the force of gravity. It is a fact of Nature, only fully explained by Albert Einstein, that gravitational forces act through the CG of an object, just like inertia. This fact can be explained at deeper levels, but such an explanation would take us too far off the subject of weight transfer. is the lift force exerted by the ground on the front tire, and is the lift force on the rear tire. These lift forces are as real as the ones that keep an airplane in the air, and they keep the car from falling through the ground to the center of the Earth. We don't often notice the forces that the ground exerts on objects because they are so ordinary, but they are at the essence of car dynamics. The reason is that the magnitude of these forces determine the ability of a tire to stick, and imbalances between the front and rear lift forces account for understeer and oversteer. The figure only shows forces on the car, not forces on the ground and the CG of the Earth. Newton's third law requires that these equal and opposite forces exist, but we are only concerned about how the ground and the Earth's gravity affect the car. If the car were standing still or coasting, and its weight distribution were 5050, then would be the same as . It is always the case that plus equals , the weight of the car. Why? Because of Newton's first law. The car is not changing its motion in the vertical direction, at least as long as it doesn't get airborne, so the total sum of all forces in the vertical direction must be zero. points down and counteracts the sum of and , which point up. Braking causes to be greater than . Literally, the ``rear end gets light,'' as one often hears racers say. Consider the front and rear braking forces, and , in the diagram. They push backwards on the tires, which push on the wheels, which push on the suspension parts, which push on the rest of the car, slowing it down. But these forces are acting at ground level, not at the level of the CG. The braking forces are indirectly slowing down the car by pushing at ground level, while the inertia of the car is `trying' to keep it moving forward as a unit at the CG level. The braking forces create a rotating tendency, or torque, about the CG. Imagine pulling a table cloth out from under some glasses and candelabra. These objects would have a tendency to tip or rotate over, and the tendency is greater for taller objects and is greater the harder you pull on the cloth. The rotational tendency of a car under braking is due to identical physics. The braking torque acts in such a way as to put the car up on its nose. Since the car does not actually go up on its nose (we hope), some other forces must be counteracting that tendency, by Newton's first law. cannot be doing it since it passes right through the cetner of gravity. The only forces that can counteract that tendency are the lift forces, and the only way they can do so is for to become greater than . Literally, the ground pushes up harder on the front tires during braking to try to keep the car from tipping forward. By how much does exceed ? The braking torque is proportional to the sum of the braking forces and to the height of the CG. Let's say that height is 20 inches. The counterbalancing torque resisting the braking torque is proportional to and half the wheelbase (in a car with 5050 weight distribution), minus times half the wheelbase since is helping the braking forces upend the car. has a lot of work to do: it must resist the torques of both the braking forces and the lift on the rear tires. Let's say the wheelbase is 100 inches. Since we are braking at one g, the braking forces equal , say, 3200 pounds. All this is summarized in the following equations: With the help of a little algebra, we can find out that Thus, by braking at one g in our example car, we add 640 pounds of load to the front tires and take 640 pounds off the rears! This is very pronounced weight transfer. By doing a similar analysis for a more general car with CG height of , wheelbase , weight , static weight distribution expressed as a fraction of weight in the front, and braking with force , we can show that These equations can be used to calculate weight transfer during acceleration by treating acceleration force as negative braking force. If you have acceleration figures in gees, say from a Ganalyst or other device, just multiply them by the weight of the car to get acceleration forces (Newton's second law!). Weight transfer during cornering can be analyzed in a similar way, where the track of the car replaces the wheelbase and is always 50%(unless you account for the weight of the driver). Those of you with science or engineering backgrounds may enjoy deriving these equations for yourselves. The equations for a car doing a combination of braking and cornering, as in a trail braking maneuver, are much more complicated and require some mathematical tricks to derive. 
Quote, originally posted by BigBlockBug » 
Bingo, thats all that needs be said, and as for the ignorance above about just beleiving the telemetry to be gospel rather than open your mind and look at the video http://****************.com/smile/emthdown.gif , thats what seperates a discussion from an argument. So you can argue all you want from your myopic view of what yopu think you know about cars. Why not branch out and consider an approach or idea and contribute to the share of ideas. Thats what ruins this forum too many people that just piss every one off. 