# RPM calculated from forward speed and wheel dia.?

A vehicle with a wheel 26 inches in dia. traveling at 10 mph.

What is the rpm of the wheel?

If you would include the formula that would be great.

Relevance

First of all, we want to calculate RPM (revs per minute), and we have the speed in MPH (miles per hour). So let's restate the speed in units of feet/minute:

(10 mph x 5280 feet/mile) / 60 min/hr = 880 feet/min

The diameter of the wheel is 26 inches. In one revolution, it travels a distance equal to the circumference of the wheel, which is PI times the diameter:

(26 inches x 3.14159265) / 12 inches/foot = 6.81 feet

Theen we just calculate the number of revolutions needed to travel 880 feet:

( 880 feet/min ) / ( 6.81 feet/rev ) = 129.3 RPM

Mostly this problem is just about scaling everything to consistent units.

RE:

RPM calculated from forward speed and wheel dia.?

A vehicle with a wheel 26 inches in dia. traveling at 10 mph.

What is the rpm of the wheel?

If you would include the formula that would be great.

Source(s): rpm calculated speed wheel dia: https://biturl.im/3m7J4
• Anonymous
5 years ago

For the best answers, search on this site https://shorturl.im/axdoT

Okay, first figure the tire diameter. The way to do this is shown below. (first link) The standard tire for that car is a P205/75 R15. That is, it's a 15" rim 205mm wide with a 75% aspect. So, first the section height: .... 205mm * .75 = 153.75mm = 6" Now, the diameter: ... 2*section + rim = (6" * 2) + 15" = 27" Now, the circumference: ... pi * diameter = pi * 27" = 84.8" = 7.07 ft. To figure RPM it'll be easiest to compute ft. per min @ 50mph.: ... 50 m/h * (5280ft/mi) * (1 hr / 60 min) = 4400 ft/min And finally: ... (4400 ft/min) * (1 rev. / 7.07 ft) = 622 rev/min. Side note... that Javascript calc appears very broken. The figures it gives are way off. Another side note... a 18" diameter would be that for a 13" rim, maybe ... way too small for a Caravan. And here's a sanity check: work backwards from the axle ratio. For that vehicle this is anywhere from 2.37 to 3.91, depending on year, engine, trim level, etc. (second link) Let's assume say 3.19 for the '98 model with the 3.0L engine. That'd work out to be 1984 RPM at 50mph (at 622 wheel RPM) in the 1:1 top gear (no slippage - even back in '98 they had lockup converters), or 2381 RPM at 60. I'd say that's about right for that engine.

• Anonymous
5 years ago

Here is a simple equation you can use which assumes that the wheel has no slip, for your case it virtually does not so this equation works fine. v=r*w, vehicle velocity = tire radius*tire angular velocity. go measure the distance from the center of the tire to the outside of the tire, it will probably be between 15"-20". To solve for wheel angular velocity rearrange the equation to get w=v/r make sure your units are consistent, i.e velocity in ft/s radius in ft. Answer will be in rad/s which can be converted to rpm, 2*pi rad/1 rev. 60sec/min etc.

• 10 mi x 5280 feet/mi x 12 in/foot = 633600 inches per hour

Since the circumference of a wheel is 2*pi*radius the wheel moves 2*3.1415*13 or 81.679 inches per revolution.

633600 inches per hour divided by 81.679 inches per revolution = 7757.1958 revolutions per hour or 129 rpm

• Anonymous