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MECHANICAL FORMULAS PROBLEM

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10 years 9 months before - 10 years 9 months before #47310 by manual-mecanica
Mechanical formulas published by Manual-Mecanica
Hello good, I have a problem but I lack formulas to solve it. I wanted to know if anyone could help me, the problem is as follows:

a car goes to 1,500 rpm, if it goes at 3rd speed (3rd ratio = 1.4) and has a differential group with a 25/40 relationship and as well as an epicloid reduction 1/3. What speed do the wheels turn?

I have no more information and that makes me very much spin, because then ajor is the simplest thing in the world

a greeting
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10 years 9 months before #47313 by manual-mecanica
Manual-Mecanica response on the problem with mechanical formulas
The data you need would have to mount the vehicle in a Dyno there connect it with several sensors and get all the data until its braking I do not know what country you are but not everyone has said device greetings

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10 years 9 months before - 10 years 9 months before #47324 by manual-mecanica
Manual-Mecanica response on the problem with mechanical formulas
Hello, I have not done a problem of these years, but I think you just have to make rules of three, but the result will give you in RPM. If you want in km/h you need to know the tire diameter.
1500rpm: 1.4 = 1071.42Rpm output in change.
1071.42RPM x 25: 40 = 669.64RPM output in the differential.
669.64rpm: 3 = 223.21rpm on the wheel.


To know in km/h, if for example you mount a 195/60 R15 wheel, you must first know the total diameter of the tire, then the distance it travels in


1

return and
final 117mm .




​1000000
=
0.00193km
.
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10 years 9 months before #47328 by manual-mecanica
Manual-Mecanica response on the problem with mechanical formulas

Maurog wrote: The data you need would have to mount the vehicle in a Dyno there they connect it with several sensors and get all the data to the braking of the same I do not know what country you are but not everyone has said device greetings

Thanks Maurog for answering me, but it's just a written problem, it's not for any car.

Lord Strange wrote: Hello, I have not done a problem of these years, but I think you just have to make rules of three, but the result will give you in rpm. If you want in km/h you need to know the tire diameter.
1500rpm: 1.4 = 1071.42Rpm output in change.
1071.42RPM x 25: 40 = 669.64RPM output in the differential.
669.64rpm: 3 = 223.21rpm on the wheel.


To know in km/h, if for example you mount a 195/60 R15 wheel, you must first know the total diameter of the tire, then the distance it travels in


1

return and
final 117mm .




​1000000
=
0.00193km
.

Hello Lord Strange, the first thanks for the answer. You will see myself realized that the problem was wrong. They ask me for angular speed of the wheels.

I know there is a formula for this, I have to use the RPM of the wheel, I suppose, right?
And the result remains in radianes/seconds, right?

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10 years 9 months before #47335 by manual-mecanica
Manual-Mecanica response on the problem with mechanical formulas
Hello, to convert revolutions per minute to radianes per second you can use the conversion factor 1rad/seg = 9.55rpm or the formula (which should be what you were looking for) w = 2 x pi x f.
w = angular velocity
2pi = 1 revolution (remember that 1revolution (360º) are 2pi radians)
F = rotary speed (number of revolutions)

Therefore, if you had 223.21rpm in the wheel

w = 2 x 3,1416 x 223.21 = 1402.47Radianes per minute.

If the problem asks for the result in radians per second divides the result between 60
1402.47: 60 = 23.37 rad/sec.

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