Unit 3 - Problem 5 => San Francisco Parking

Given:

• Friction between tires and road prevents car from slipping.
• $F_f$ has a max value of $80\%$ of car weight.
• The maximum angle is $\alpha$.

Reasoning

• $F_{max} = 0.80 * W_{car}$. Beyond $F_{max}$ the car slides down the hill.
• $F_{parallel} = W_{car} * \sin\alpha$ (obtained from previous problem)is the actual force on the roadway.
• The maximum angle occurs when the $W_{car} * \sin\alpha = 0.80 * W_{car}$

Calculations:

$$\sin\alpha = \left(\frac{.8W_{car}}{W_{car}}\right)$$
$$\sin\alpha = .8$$
$$\alpha = \arcsin{.8} = 53.13^{\circ}$$

 

Some observaions

 

I do not ever want to be on a grade of 53%. I could not stand up on it. Consider the following:

• The famous Lombard Street in San Francisco, the curviest street in the world only has a grade of 14.3 %.
• The steepest street in San Francisco is Filbert Street between Hyde and Leavenworth at a grade of 31.5 %.
• An interstate is out of standard if it has a grade > 7%.
• Local roads are much higher. 12% or 15% are sometimes allowed.
• Driveways can be as much as 30% for a short distance
• Railroads generally avoid grades over 2%.
• Even commercial airplanes don't have numbers as high as 53%,
• A330, out of maintenance - empty, engineers asked for a full power takeoff. Full back stick after normal rotation to stop speed going silly, settled at about 28 degrees, 8,500 f.p.m climb
• 747 stick shaker starts at 17-17.5 deg, so we never intentionally go above that.

The concept of slope (slope is normally described by the ratio of the "rise" divided by the "run" between two points on a line. ) applies directly to grades or gradients in geography and civil engineering. Through trigonometry, the grade m of a road is related to its angle of incline theta.

$${incline} = \tan\theta$$ $$\theta = \arctan{(incline)}$$