Design Monorail: Gear Ratio Calculation for Climb 45° Slope

[Pfloyd77]

I have to build a monorail that can climb a 45 degree slope length of the track is 8m. The mass of the monorail is .3kg and it has to carry a mass of 1kg up the slope.im given a 15V electric dc motor with:

Power = 6W at motor stall
no load rpm at 0.07 Amp = 3800
stall torque at 0.4 amp = 0.01N

i calculated that the force needed for the monorail to overcome static friction is 17.2N

i have to use the above info to determine what my gear ratio should be so that the monorail can travel up the slope...does anyone know how to do this?

 

[Danger]

Welcome to PF, Pfloyd. I'm afraid that I can't help you with that. I'm more of a 'trial and error' designer with a penchant for overkill. I'd be trying to shoehorn a Hemi into it.

 

[mgb_phys]

Calculate the force needed along the track (friction + component of gravity)
Then work out what wheel diameter would give you the limiting torque for this force.
Then you have the ratio of the shaft diamter to the wheel diamter.

 

[Mech_Engineer]

Sounds to me like you have all the information you need to solve the problem.

Perhaps you should try doing a free body diagram of the monorail to quantify all of the forces acting on it; that will give you a net force acting on your vehicle, which can then be used to decide exactly how much gear reduction you will need.

 

[Pfloyd77]

Calculate the force needed along the track (friction + component of gravity)
Then work out what wheel diameter would give you the limiting torque for this force.
Then you have the ratio of the shaft diamter to the wheel diamter.

how do go i go about doing step 2 using all of my motor values

we never covered any motor equations in class so I am really new at the motor equations and knowing how to apply them

 

[Pfloyd77]

Sounds to me like you have all the information you need to solve the problem.

Perhaps you should try doing a free body diagram of the monorail to quantify all of the forces acting on it; that will give you a net force acting on your vehicle, which can then be used to decide exactly how much gear reduction you will need.

after I've calculated my net force how do i compare that with the force that is generated by my motor...what formula(s) would i need to use

 

[Mech_Engineer]

Well, mgb and I have assumed you will be driving this monorail with some kind of wheel; so by taking the diameter of the wheel and multiplying it by the force you will need to overcome gravity and friction you can get the torque required to move the system.

You have a stall torque rating for the motor and a torque required, so figure out how many time you need to multiply the motor's torque to get the required torque. There's your gear box ratio.

 

[mgb_phys]

compare that with the force that is generated by my motor

There isn't a force generated by your motor there is a torque = a force at right angles to a distance. The 0.01 N should probably be 0.01Nm (check the specs)

So you know the motor can lift 0.01N at 1m radius, then what radius can it generate the force you have calculated?

 

[Pfloyd77]

There isn't a force generated by your motor there is a torque = a force at right angles to a distance. The 0.01 N should probably be 0.01Nm (check the specs)

So you know the motor can lift 0.01N at 1m radius, then what radius can it generate the force you have calculated?

the force i calculated was 17.72N

i got this answer by using the following
m=1.3kg
g=9.8m/s^2
theta = 45
coefficient of static friction (us)= 0.9
N=mgsin45


F = (us)N +mgcos45
= (0.9)(1.3)(9,8)(sin45) +(1.3)(9.8)(cos45)
F=17.71

since the motor can lift 0.01N at a 1m radius than i would need a radius of 1,772m??
that answer seems way too large did i do something wrong?

 

[mgb_phys]

You have done your sums wrong.
If the slope is at 45deg then the normal force and the force downward should be the same. Your motor has to overcome both the weight and the friction.

You know about torque ? If you can lift 0.01N at 1m what can you lift at 2m?
Try this with a heavy book and an outstretched arm!

 

[Pfloyd77]

You have done your sums wrong.
If the slope is at 45deg then the normal force and the force downward should be the same. Your motor has to overcome both the weight and the friction.

You know about torque ? If you can lift 0.01N at 1m what can you lift at 2m?
Try this with a heavy book and an outstretched arm!

for my sums i made a new axis in my fbd that coincides with both N and F(friction) where the normal is equal to the force of fravity (mgsintheta) i don't understand how i have done my sums wrong

as for the 2nd part i would need 1/1,772 m

 

[mgb_phys]

Sorry I misunderstood, the 17.71 is both to overcome friction and the component of the weight - the OP doens't list the coeff of friction.

So if you have 17.71N of force to provide and only 0.01Nm of torque then you need a wheel diameter of 0.01Nm/17.71N = 0.5mm since this is a bit unlikely your are going to need some gears.
If the shaft diamter of the motor is say 5mm then you need a 10:1 gear box (assuming no losses)

 

[Pfloyd77]

Sorry I misunderstood, the 17.71 is both to overcome friction and the component of the weight - the OP doens't list the coeff of friction.

So if you have 17.71N of force to provide and only 0.01Nm of torque then you need a wheel diameter of 0.01Nm/17.71N = 0.5mm since this is a bit unlikely your are going to need some gears.
If the shaft diamter of the motor is say 5mm then you need a 10:1 gear box (assuming no losses)

my actual shaft diameter is 2mm which would give me a gearbox ratio of of 4?

 

[mgb_phys]

I'm not sure of your exact definition of gearbox ratio but I would say yes.

 

[Mech_Engineer]

my actual shaft diameter is 2mm which would give me a gearbox ratio of of 4?

Pfloyd77, you still haven't told us exactly HOW the monorail is driven. How is the motor's power put down to the track? A gear? A wheel? A pully? How are you going to make the monorail move?