This is evolution of the Braking Geometry thread. George M. did such a nice job with the ghost, it deserves a new thread.

Is this your final answer?

Interesting.

Looks like the center of gravity may be too high. I would guess about 25 to 30". At least for this car, an encloses sedan would be higher then a roadster.

Jim

I have not taken any measurement or ran any calculations. But I see two possible flaws in the calculation. But I only took a quick glance at the diagram. I would want to know how the center of gravity is/was determined. I would guess your figures would indicate that the CG is at the top leading edge of the seat. I would say CG would greatly very based on the body style of the car. For example a fordor with the extra steel and glass would have a CG is much higher than an open touring. The other major issue I see; is the vehicle design, with its inherent weight distribution. I will base this on my 26 roadster; if you eliminate any thought of the engine for a moment. I would say the weight is fairly distributed across the length of the vehicle evenly. With that said I would say placing CG at the leading edge of the seat is about correct. I would guess CG would then be a few inches above the floor (again thinking about a roadster, and not a closed car). Now let’s talk engine. I have never weighed one, but I have heard they weigh about 500 pounds. That maybe a little high but it is a nice easy number. If that is correct the engine’s weight is 1/3 of the total weight of the roadster. Do to its mass it is not a trivial component in the overall CG calculations. Now due to the transmission hanging of the engine, I would say the CG of the engine is inline of the 3rd cylinder, about 2 inches or so from the bottom of the block, (again I have not looked at the car or taken any measurements). If that estimate is correct, the height would be about the same as the rest of the vehicle. BUT; now you need to calculate the two CGs, The back of the napkin method, I would say connect the two dots and you will be about 1/3 the way forwards from the leading edge of the seat. Where does that put us? I would say CG is located about where the peddles come through the floor.

Now with that info, I would say your CG location is wrong. Now with 2 adult passengers, in the front seat. You might have the correct CG location. Then you need to adjust for the additional load (+300 to 400 lbs), again in a T this weight is a large portion of the total load.

I just entered your formulas into excel used your values. I do not think I could stop my T in 84 ft from 30mph. I would say she is well tuned. Maybe in an emergency hard stomp on the brakes I could, but I would not want to cause any damage to the T just to try it. Not sure why but my calculations are a bit higher than yours. Also I would believe a 180 ft stopping distance at 30 mph. 84 ft is just way to fast. I used your formulas but my values are much higher than yours.

Thanks for the inputs, guys. I chose 40" cg height because it was convenient with the grid I was using, and mid-point of wheelbase for ease. Jersey George supplied the ghost later. I agree, the cg would be lower on that car, and also further forward. This is just a starting point to understand the dynamics of braking in a T.

I assumed newish tires with coefficient of friction of 1.0. Tires harden with age and lose traction.

To get stopping distance, I first had to calculate weight shift due to braking, and hence, g force. I plugged in numbers until I came close to braking force equaling weight on rear wheels, which would be a c/f of 1.0, of course. There's no doubt a handy formula, but cut and try worked in this case.

Stopping Distance

I inserted some of my own numbers in the following for continuity:

My Notes:
W has been changed to G in the formulas to conform to other formulas used.
* is multiplication symbol; / is division.
v is velocity in mph
g is gravity, 32.2 feet per second per second
-------------

Taken from 1909 Audel's "Self-Propelled Vehicles."

1 mile per hour = 1.466 feet per second.

Whence G*v*v/2g = G*(1.466*1.466)/64.4 = G*2.15/64.4 = G*0.0334

Then a vehicle weighing 1200 lb, traveling at ten and twenty miles per hour, by the formula,

K = G*v*v*0.0334,

in which V represents miles per hour, will be for 10 mph: 1200*100*0.0334 = 7,480 foot pounds;
for 20 mph: 1200 X 400 X 0.0334 = 29,920 foot pounds.


To Find Distance in Which Brakes Will Act on Vehicle's Speed.

Then, taking k as the coefficient of friction between'the tires and road surface, which is approximately 0.60 for rubber tires; and taking w as the proportion of the total weight carried by the wheels to which the brake is applied, which may be assumed to be 0.6 of the whole, the maximum distance required to stop the vehicle on the level, on an ordinary road, whose surface resistance is, supposedly, included in the expression, k, may be expressed by 1, as follows:

S = G*v*v*O.0334/kw

Then, for a vehicle weighing 1200 lbs, tired with average rubber tires, traveling at a momentum of 10 and 20 miles per hour, respectively, we have:

S = 4008/0.6*720 = 9.3 feet at 10 miles, and
S = 16032/0.6*720 = 37.1 feet at 20 miles;

these distances representing the maximum, with a braking effect sufficient to cause the wheels to skid.

----------------

You might try the formulas again, Jason.

rdr

Oh, boy. I see I'm using w for two different things. Rev. 3 is pending...

Rev. 3:

Hmm, had to revise my Audel translation also:

Taken from 1909 Audel's "Self-Propelled Vehicles."

W has been changed to G in the formulas to conform to other formulas used.
* is multiplication symbol; / is division.
v is velocity in mph
g is gravity, 32.2 feet per second per second
p is pounds of weight on braking wheels

1 mile per hour = 1.466 feet per second.

Whence G*v*v/2g = G*(1.466*1.466)/64.4 = G*2.15/64.4 = G*0.0334

Then a vehicle weighing 1200 lb, traveling at ten and twenty miles per hour, by the formula,

K = G*v*v*0.0334,

in which V represents miles per hour, will be for 10 mph: 1200*100*0.0334 = 4,008 foot pounds;
for 20 mph: 1200 X 400 X 0.0334 = 16,032 foot pounds.


To Find Distance in Which Brakes Will Act on Vehicle's Speed.

Then, taking k as the coefficient of friction between'the tires and road surface, which is

approximately 0.60 for rubber tires; and taking p as the proportion of the total weight carried by

the wheels to which the brake is applied, which may be assumed to be 0.6 of the whole, the maximum

distance required to stop the vehicle on the level, on an ordinary road, whose surface resistance

is, supposedly, included in the expression, k, may be expressed by 1, as follows:

S = G*v*v*O.0334/kp

Then, for a vehicle weighing 1200 lbs, tired with average rubber tires, traveling at a momentum of

10 and 20 miles per hour, respectively, we have:

S = 4008/0.6*720 = 9.3 feet at 10 miles, and
S = 16032/0.6*720 = 37.1 feet at 20 miles;

these distances representing the maximum, with a braking effect sufficient to cause the wheels to
skid.