I know the drag force is 1108N (113kg). But if I understand you correctly center of vertical forces has no effect on center of horizontal forces? Now that you have net vertical force and drag how do you calculate this? Isn't horizontal forces center just a halfway distance between floor and roof...
Yes. What I meant to say is that the long horizontal line represents aero torque measured from the rear axle. The start of the long horizontal line represents center of vertical aero forces obviously. It doesn't matter that it is outside the car body dimensions. It was calculated from the above...
No I mean the start of the longest line below the car from the left. We are discussing center of aerodynamic forces or aerodynamic pressure. It will be expressed as a position in meters from the rear axle. As you can see I already have a position for a horizontal axis, but don't know how to set...
I am bumping this thread to follow up the discussion with a different question. So we have determined center of horizontal forces but what about vertical center? Look at the picture below. The start of the long horizontal line represents center of horizontal aero forces. How does one then...
Yes I meant to wrote up. OK. So if we take this method of calculating a force there is one problem. What happens if there was equal amount of forces acting in opposite directions like 50kg lift at the front, -50kg rear negative lift a car would have a large understeer in reality. But in a...
OK. So you are saying 54kg-26kg = 28kg
Then 54kg/28kg = 1,93 or 193% of the wheelbase. The result being 4,709m to the left of the right axle.
So a force equal of 28kg is pressing down 4,709m to the left of the right axle. Is this correct?
If both forces acted in the same direction I would add 54kg + 26kg = 80kg
Then 54kg/80kg = 68% front aero balance. Then multiply wheelbase with 0,68.
0,68 x 2,44m = 1,66m meaning center of forces being this distance away from rear axle.
But I don't know how to calculate if forces are pointed...
Hello. I am working on a physics project for a simulation title and have stumbled upon on an interesting challenge.
Below is the example from wind tunnel data of a Dodge Viper GTS sports car.
Wheelbase: 2,44m
Lift front axle: 54kg
Negative lift rear axle: 26kg
Can somebody please explain to...