what is the circumference of a bus wheel/tyre? I NEED HELP ASAP!!?
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Measure the diameter and multiply by pi (3.1415)
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15a) Ahh, lengths, areas, and volumes. Lengths are proportional to each other, areas are proportionatl to the square of the length, and volumes are proportional to the cube of the length.\n\na) The ratio of the area of the windscreen is ⅜^2\nb) The ratio of the capacities of their boots is ⅜^3\nc) The ratio of the widths is ⅜\nd) I suppose the model car has four wheels, just the original. It's a model, after all.\n\n15b) we are looking for a rectangle with area equal to 16. Let the width be w. Then the length is w + 6.\n\n16 = w(w + 6)\nw^2 + 6w -16 = 0\n(w + 8)(w - 2) = 0\n\nw = 2, and the length is 8. The area is 16.
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u dont need the 6 in this question because all the tyres travel the same amount. (its juss there 2 throw u)\nso then all we have to do is go:\n\n 16000(distance travelled) / (tht means divide)8 (amount of wheels)= 2000 (basic division)
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I get the radial acceleration = ar = V²/r = (325*1000/3600)²/194 = 42.01 m/sec²\nI agree with your value for at = 6.69 m/sec²\n\nThe total acceleration = A = √(42.01² + 6.69²) = 42.539 m/sec²\n\nThe Cf required = A/g = 42.539/9.8 = 4.34\n\nTypical tire/asphalt Cf is around 1.0\n\nAmazing what a little bit of bank angle can do for you!
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#30 is 58.88.  Area = pie*(diameter/2) squared\n#38 is 900.  Area of triangle = .5 base*height\n\n#44 is 73,950\n\nCircumference = Pi X Diameter\n\nSo 3.14 X 2.33 ft. (14 in tireX 2 to convert to diameter)/(12 to convert to feet)=7.14 feet per tire revolution and the trip is 100 miles (5280 ftx100) or 528,000 ft.  528,000 ft/7.14= 73,950 tire rotations to complete the trip.
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Go to www.tirerack.com and enter your vehicles info (or call their 800 #). They will know.
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Manhole covers must be a minimum of 22 in (56 cm) in diameter, but can be as much as 60 in (1.5 m) in diameter.  Without knowing the specific diameter, we won't be able to give you the circumference.\n\nIf you do have a diameter, though, the circumference is:\nC = pi * diameter.\n\nUsing the numbers above, the circumference of a manhole cover will be between 69 inches and 189 inches.
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The first number ( "26" or "700" or "27") indicates the tire and wheel diameter. The second number ("1.5," "2.0," "25c," and "1 1/4") refers to the width of the tire as in 1.5 inches; 2.0 inches; 25mm; and 1 1/4 inches; and so on.\nDon't be confused by the road bike tire designation "c" as in the size 700 x 25c. The "c" is just a French tire size, it doesn't stand for a numerical measurement.
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Please check out this web site-it should be very helpful:\n\nhttp://www.maxxis.com/Services/how_a_tire_is_made.asp
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circumference = pie times diameter\nradius is half of the diameter\n\nso if you have either the circumference or the diameter, you can figure it out.
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Question 1:\n\nTorque is a force applied at a distance,\nT = F * r\nWhere T is the torque, F is the force, and r is the radius the force is applied away from the center or rotation.\n\nT = 40 Newtons * .15 meters = 6 Newton meters\n\nA torque causes an angular acceration,\nT = alpha * I\nwhere T is the torque,  alpha is the angular acceleration, and I is the rotational inertia ("moment of inertia").\n\nAn angular acceleration relates to a linear acceleration,\nalpha = a / R\nwhere alpha is the angular acceleration, a is the linear acceleration, and R is the radius of rotation (the wheel's radius in this case).\n\nThe linear acceleration can be found by finding the change in linear velocity with respect to time,\na = delta v / delta t\na = 2.25 m/s / 3 s = .75 m/s^2\n\nalpha = .75 m/s^2 / .5 meters = 1.5 radians per second^2\n\nT = alpha * I \n--> I = T / alpha\nI = 6 Nm / 1.5 rad/s^2 = 4 kg m^2\n \nDo your own homework next time (like on question 2).\nhttp://sln.fi.edu/qa97/spotlight3/spotlight3.html
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b = 70sin55\n\nb = 57.34
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So you need to know the diameter of the rotating thing. Then it is simple dimensional analysis. Let's say the wheel rotating is 1m in diameter. So you have:\n\nR/M  and you want   m/s\n\nOk M->s is easy, you multiply by 60 (or in this case 1/60):\n\nR/(60M) = R/s\n\nTo change R to m, you need to find the circumfrance of the circle in meters. Which is pi * the diameter. So 3.141 * 1 = 3.141 meters. So each rev is 3.141 meters so multiply by that.\n\n3.141R/(60M) = m/s\n\nGood. So now you have a formula:\n\n[(3.141 * d)/60]*(R/M) = m/s
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Divide 34 by PI...that will give you the diameter.\n\nDiameter = Circumference / PI
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it's approximately 6 m.\n\nthe formula that relates diameter and circumference is:\n\nCircumference = pi * diameter\n\n18.85 = 3.14 * d\n\n18.85/3.14 = d\n\n6.00 = d  <-- answer
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He is right. Those measurements are for something else. Bearings are standard size and should fit any skateboard wheels, provided that you have skateboard bearings and skateboard wheels. :) It's all preference.
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It is unlikely that you will have any measurement on the drawing that is related to a radius by pi. You double the radius to get the diameter, and multiply that by pi to get the circumference, but you are very unlikely to get a drawing in which the circumference is given, but not the radius. The radius is the gap you would set on your compasses to draw the circle. If you did need to design a wheel of a given circumference, maybe for a gearbox,  then you would divide by 2pi to get the radius needed.