Great Cars- OLDEST AUTOS

The mathamatical formulas didn't translate so well where 1/2 or 1 OVER 2, appears as 12cPdAv2.... But you can ask the same questions, and copy ALL and past the entire document into a competent modern word processing program.

https://chatgpt.com/

what is the area of a circle 4" in diameter?
ChatGPT said:
ChatGPT

To calculate the area of a circle, you can use the formula:
A=πr2
A=πr2

Where:

AA is the area of the circle.
rr is the radius of the circle.
ππ is approximately 3.14159.

Step 1: Convert diameter to radius

The diameter of the circle is given as 4 inches. The radius (rr) is half of the diameter:
r=42=2 inches
r=24​=2inches
Step 2: Calculate the area

Now, using the formula for area:
A=π×(2)2=3.14159×4=12.56636 square inches
A=π×(2)2=3.14159×4=12.56636square inches
Final Answer:

The area of the circle with a 4-inch diameter is approximately 12.57 square inches.
You said:
12.6"2 x 30PSI/I2 = how many pounds of force in total
ChatGPT said:
ChatGPT

To calculate the total force, we can use the formula that relates pressure, area, and force:
Force=Pressure×Area
Force=Pressure×Area

Where:

Pressure is given as 30 PSI30PSI (pounds per square inch).
Area is given as 12.6 in212.6in2.

Step 1: Multiply Pressure by Area
Force=30 PSI×12.6 in2
Force=30PSI×12.6in2
Force=378 pounds of force
Force=378pounds of force
Final Answer:

The total force is 378 pounds of force.
You said:
How many Kn per square meter is 378 pounds of force over a square meter?
ChatGPT said:
ChatGPT

To convert pounds of force over a square meter (lbf/m²) to kilonewtons per square meter (kN/m²), we need to use the conversion factor between pounds-force (lbf) and newtons (N).
Conversion Factors:

1 pound-force (lbf) = 4.44822 newtons (N)
1 kilonewton (kN) = 1000 newtons (N)

Step 1: Convert 378 pounds of force to newtons

First, we convert 378 pounds of force to newtons using the conversion factor:
Force in newtons=378 lbf×4.44822 Nlbf
Force in newtons=378lbf×4.44822lbfN​
Force in newtons=1688.7 N
Force in newtons=1688.7N
Step 2: Convert newtons to kilonewtons

Now, convert the force in newtons to kilonewtons by dividing by 1000:
Force in kilonewtons=1688.71000=1.6887 kN
Force in kilonewtons=10001688.7​=1.6887kN
Final Answer:

The total force of 378 pounds of force over 1 square meter is approximately 1.689 kN/m².

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This is a nice sojourn through the history of the auto-mobubbile.

The mathamatical formulas didn't translate so well where 1/2 or 1 OVER 2, appears as 12cPdAv2.... But you can ask the same questions, and copy ALL and past the entire document into a competent modern word processing program.

https://chatgpt.com/



Assuming a standard or typical sedan with a frontal surface area of 2 meters wide and 1.5 meters high, how much power if required to overcome the air resistance at 120 Kmh?
ChatGPT said:
ChatGPT

To calculate the power required to overcome air resistance for a typical sedan at a speed of 120 km/h, we need to use the equation for aerodynamic drag force and then determine the power based on that drag force. The equation for aerodynamic drag force is:
Fd=12ρCdAv2
Fd​=21​ρCd​Av2

Where:

FdFd​ = drag force (in Newtons, N)
ρρ = air density (approximately 1.225 kg/m³ at sea level and 15°C)
CdCd​ = drag coefficient (typically 0.3 for a sedan)
AA = frontal surface area (in square meters, m²)
vv = speed (in meters per second, m/s)

Step 1: Convert speed to meters per second

The speed given is 120 km/h. We need to convert it to meters per second:
v=120 km/h=120×10003600=33.33 m/s
v=120km/h=3600120×1000​=33.33m/s
Step 2: Calculate the drag force

We assume the following values:

ρ=1.225 kg/m3ρ=1.225kg/m3 (air density)
Cd=0.3Cd​=0.3 (typical for a sedan)
A=2 m×1.5 m=3 m2A=2m×1.5m=3m2
v=33.33 m/sv=33.33m/s

Now, plug these values into the drag force equation:
Fd=12×1.225×0.3×3×(33.33)2
Fd​=21​×1.225×0.3×3×(33.33)2
Fd=0.6125×3×1111.09
Fd​=0.6125×3×1111.09
Fd=0.6125×3333.27=2043.6 N
Fd​=0.6125×3333.27=2043.6N
Step 3: Calculate the power required

The power required to overcome the drag force is given by:
P=Fd×v
P=Fd​×v

Substitute the values for FdFd​ and vv:
P=2043.6 N×33.33 m/s
P=2043.6N×33.33m/s
P=68,134.7 W or 68.1 kW
P=68,134.7Wor68.1kW
Final Answer:

The power required to overcome air resistance at 120 km/h for this typical sedan is approximately 68.1 kW.
You said:
What is the pressure on the front of this car, in Kn/m2
ChatGPT said:
ChatGPT

To calculate the pressure on the front of the car due to aerodynamic drag, we can use the following equation for the dynamic pressure (qq):
q=12ρv2
q=21​ρv2

Where:

qq = dynamic pressure (in Newtons per square meter, N/m² or Pascals, Pa)
ρρ = air density (typically 1.225 kg/m³ at sea level and 15°C)
vv = speed of the vehicle (in meters per second, m/s)

Step 1: Convert speed to meters per second

The speed of the car is 120 km/h, which we already converted earlier:
v=120 km/h=33.33 m/s
v=120km/h=33.33m/s
Step 2: Apply the dynamic pressure equation

Now, we use the dynamic pressure formula with:

ρ=1.225 kg/m3ρ=1.225kg/m3
v=33.33 m/sv=33.33m/s

Substitute these values into the equation for qq:
q=12×1.225×(33.33)2
q=21​×1.225×(33.33)2
q=0.6125×1111.09
q=0.6125×1111.09
q=679.3 Pa
q=679.3Pa
Step 3: Convert Pascals to kilonewtons per square meter

Since 1 Pascal (Pa) is equal to 1 N/m², we can convert the result to kN/m² by dividing by 1,000:
q=679.31000=0.679 kN/m2
q=1000679.3​=0.679kN/m2
Final Answer:

The pressure on the front of the car due to aerodynamic drag is approximately 0.679 kN/m².

Chat GPT is I think, wrong in the final calculation but this gives me the ratio for thrust on a piston face, at 30 PSI - all the stroke, as a ratio of the frontal surface area of the car.