Sewer manhole lid worth saving?

RJ, that’s a good analysis, but 30 MPH is 44 ft/sec, so the car takes 2/44 ths of a second, or about 45 milliseconds.
That changes the conclusion quite a bit.
I misplaced a decimal point in the post. The time is 45 msec. the vertical drop is .396". I had rounded up the time to 50 msec. which gave a drop of .48", rounded to .5". The conclusion was OK, lousy transcription.
 
Ha ha, I stopped doing the math just one step too soon. Ah me.
 
.396"? About like hitting a 3/8" sheet of plywood lying on the road. Tire deformation should help absorb some of that. Doesn't sound very drastic.

Extrapolating RJ's analysis, I think you can easily assume that a wheel (aka unsprung mass) only masses about 1/10 of the amount of mass that it is supporting, or less. Therefore the spring is compressed with a force about 10 times the weight of the wheel, and the wheel is going to experience an acceleration force about 10x the typical 9.8 m/s^2 ( 32 ft/sec^2). So your 1/2 a t ^2 now goes to 160 * t^2, or 3.9". I'm guessing that's closer to the real wheel vertical displacement, although the shock absorber would reduce that somewhat.

And then we'd have to analyze how much the car moves back up to know what kind of jolt the driver experienced. All we need is the Fourier model of the suspension and tire, and model the impact as a step function ....


But, I think this quote "covers" it (pardon the pun)

It's best to avoid open manholes altogether.
 
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Only if there is a supporting lip......
 
Back in the days when I used to test my shocks by bouncing the car, my recollection was that the period of the oscillation was on the order of 1 second. If one assumes that the spring is carrying a load of 1,000 lbs. above the preload created by the suspension restraint and that the spring is further compressed 6" due to that load, the spring constant would be greater than 170 lb/in. When the wheel is no longer constrained by the pavement, the concerned mass is that of the wheel and associated suspension components. 100 lbs is a fairly good guess.

The period of oscillation of the suspension would be equal to 2π(m/k)^1/2 or around .77sec. When the wheel drops, it is going through less than 1/4 of a full period or less than .2 sec. Gravity would play a minor role in the movement because of the overwhelming spring force. The wheel would have to make contact with the pavement again significantly sooner than .2 sec. in order to not have the bone jarring shock on contact.

Shock absorbers offer little restriction on extension and so would not contribute to the downward movement. When the wheel was forced upward again by the pavement, they would be in play essentially forcing a close coupling between the pavement and the chassis.

This is a greatly simplified analysis. Actual results could vary greatly because of the complex mechanical relationships. An actual test would be in order but I would leave that to Myth Busters.
 
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