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Originally Posted by veyron1 end since no one here, according to you, is qualified enough to talk about chassis dynamics and chassis engineering, we ARE, talking in a generalised sense... |
If you go back and read my post, you will notice that I never said any such thing. I said we are not qualified to talk about whether the chassis has "more-than-required" flex...simply because we do not know what is the "required" flex in the first place, since that data is calculated based on the same vertical loads, the same available traction that you talk about. Guess what? We dont know jack about any of the forces. I am talking specific values here.
I am not questioning your or anybody else's technical ability, I am simply saying that when you don't know the base parameters, you cannot judge the end result and say it is too much or too less.
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Originally Posted by veyron1 the chassis flex has to do with vertical loads,available traction, centre of gravity,etc. all of which result in the net cornering efficiency, that puts the chassis flex into the picture. for example, if a car has available traction of 700 units and has a vertical load of 500 units, the car's cornering efficiency stands at 140%, or the car would exert a force of 1.40 G's while cornering (theoretically), and would exert a certain amount of force on the chassis, causing it to flex. |
You may be right on whatever you said. But look at this from a holistic viewpoint. We are not looking for a stiffer chassis here. The chassis is serving its purpose pretty well. Look at the task at hand: improving the handling characteristics of the Scorpio. If you think stiffening the chassis will help, then tell us why you do.
All I said earlier was to stiffen the front and soften the rear. Now since you are a specialist in generalized vehicle dynamics and chassis engineering, pray tell me what effect a 20 N/mm increase in front spring rate and a 15 N/mm decrease in rear spring rate would have on the torsional stiffness of the chassis? Cant predict, right? Thats why we have CATIA and ANSYS.
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Originally Posted by veyron1 so, i was GENERALLY ASSUMING, that since the scorpio has a high centre of gravity, is top-heavy, and puts a lot of load on it's tyres while cornering, it has high chassis flex; and when compared to it's rivals, more so... |
Again you are comparing vehicles solely based on seat of the pants evaluations. These differences in loads and centre of gravity that you talk about with gay abandon, are more often than not talked about in grams and mm, boy you really must be tapped into the cosmos to even MAKE out such small differences....Tell me how many millions of miles might you have done in the Scorpio and its rivals to be able to make out such differences.
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Originally Posted by veyron1 if a piston is 2 inches (5.08 cm) in diameter (1-inch / 2.54 cm radius), and another piston is 6 inches (15.24 cm) in diameter (3-inch / 7.62 cm radius). The area of the two pistons is Pi * r*r. The area of the first piston is therefore 3.14, while the area of the second piston is 28.26. The second piston is nine times larger than the first piston. This means that any force applied to the first piston will come out nine times greater on the second piston, and vice versa. So, if you apply a 100-kg downward force to the first piston, a 900-kg upward force will appear on the second. The only catch is that you will have to depress the first piston 9 inches (22.86 cm) to raise the second piston 1 inch (2.54 cm).
therefore, an increase in the dia size of the master cylinder would actually increase the hydraulic pressure, with the only catch being that the piston would have to travel lesser distance; hence the smaller dia booster & a larger reservoir... |
Impressive. But consider this: The equation for force applied on the piston is -
F = (Wa/A)
where W- force exerted at the slave cylinder.
a- area of the piston
A - area of the slave cylinder.
Since A is constant in our case, the equation is F = K (Wa).
Keeping 'a' constant, if you increase F, W increases. Meaning, press harder if you want to stop faster.
Keeping W constant, if 'a' is increased, F increases. Meaning, if you install a larger diameter master cylinder, you increase pedal force for the same braking force. And if you add a SMALLER booster, all you are doing is further increasing the pedal force. All this math crap means, in effect, what I said earlier:
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Originally Posted by ananthkamath This results in spongy feel and may in fact INCREASE the braking times. |
Go read your hydraulics textbook again, this formula is given right underneath the theory that you explained to me. If you want to indulge in mathematical mud-slinging, I am all game for it. This is the stuff that I do everyday. Yet, something tells me that nothing comes of protracted debate.
We are trying to solve practical problems here, not indulging in one-upmanship as to who knows the most theory or who does the best guessing. Again I am not judging you for your theoretical knowledge or approach. I understand you want to share what you know like the rest of us here. But please refrain from making any judgments regarding technical issues. Thats even worse than judging subjective things like styling.
No offence meant or taken.
Ananth