LCP

<b>646.4</b> on our S4s. I was just thinking about the incredible rotational/centifugal G forces the tread of a tire must encounter when spinning at high velocities. The basic formula here to find rotational acceleration, if I'm not mistaken here, is acceleration equals velocity squared divided by radius of the curve.

Plugging in the numbers I find that the velocity of the rotating edge of the tire tread (which must match the vehicle's speed on the roadway) is 146.67 feet per second (100mph), and I know that the radius of one of our tires is about 1.04 feet (half of a 25" tire height). So doing the math, that yields a rotational acceleration of 20,685 feet per second per second, or dividing by 32 feet per second per second (the force of gravity, 1 G), it yields 646.4 G's. That seems like an incredible amount of centrifugal force that tire tread must withstand, and to think that if the tire radius stays the same, it varies with the square of velocity, meaning 200mph is 2585.6 G's, and a mere 10mph is still exerting 6.46 G's on the tire tread. No wonder tire blowouts are so explosive, and no wonder faster cars try to put on larger diameter tires -- to increase the radius and thereby decrease the G forces on the tire tread.

Am I right, or did I screw the physics of this statement up?

Toy_Guy

You've got it right. It is amazing. (g is 32.174 ft/s by the way)

ryans4

...i replaced the two "stock" motors (there were two becuase it ran a seperate engine/gearing for the front and the rear) with two "drag racing" motors from the hobby store.

Since the truck had open differentials (what a POOR design might I add), if you held one wheel (left to right) and hit the gas (with the other wheel in the air), the tire would begin to grow off the wheel. Pretty exciting.

LCP

IIRC, the circumference of the earth is about 25,000 miles (132,000,000 feet), and it rotates through that distance in 24 hours (86,400 seconds), so it's rotating at 1,527.78 feet per second, divided by the radius of 5,835.6 feet, yields a rotational acceleration of 0.26 feet per second squared, or about 0.008g, at the equater, which is pulling in an opposite direction as the force of gravity. So does 32.174 equal the net result of 32.43 minus 0.26?

Also what happens as you move away from the equator, and this minor force of rotational acceleration of the earth begins to occur at a diagonal angle from the true force of gravity? Is this ever noticed (i.e., does liquid in the northern hemisphere tend to flow to the south on truly flat surfaces)?

RocketBoy

any two objects. There is gravity between you and your computer. The force of attraction, F, between two bodies of mass m1 and m2 is given by :

F=G*m1*m2/r^2

where r = distance between the two bodies and G = the universal gravitational constant = 6.67 x 10-11 N m2 kg-2

Like you were suggesting though gravity varies over the entire surface of the earth based on elevation, local magnetic fields, and the makeup of the Earth's crust in that area. For instance the local gravity in Chandler, AZ where I live has been measured at 32.133526 ft/s². The local gravity in Salt Lake City is 32.147711 ft/s²

LCP

...but what I was getting at is that the rotation of the earth could make up almost a 1% difference between gravities at different latitudes of the earth, and that the rotational forces could exert force at an angle diagonal from the prevailing force of gravity. And it makes sense that gravity would be stronger the further north you go (like between AZ and UT) since rotation has less of an effect there.