Rack and Gear Design

what is needed to know is what you are driving with the gear. Resistance to the force you are applying is needed to figure your gear.

The pinion Would be driving a Gear Box with ratio 1:10, the output gear would have a 1500 RPM and then it drives a Alternator.

Is it possible to achieve the pinion size (Less than)of 12.7cm for this kind of application ?

I interpret your query to state that the rack presently exists. You say you "have" a rack.

If that is the case, then the diametral pitch of the required pinion is established. That means that the pitch diameter of the pinion is limited to certain fixed values. In other words, the pitch diameter has to include and only include a whole number of teeth.

If that is the case then it is unlikely that you can obtain both the rack speed and the pinion RPM that you have specfied. You will have to choose the closest value for the most important of the two variables.

I think that your calculations have to begin with the pitch of the rack.

Hi,

I do not have the dimention of the rack, i have just one sepcification of rack, just linear distance moved is 1 meter/sec.

It is still not clear if the rack exists or not.

I suggest that you look at the link at the bottom on this page to see what is commercially available. Go to where it says "you may be interested..."

You definitely need some help!

First, you say you want to transfer 15kW of power to (or from) this rack with a speed of 1 m/s.

Power=Force*Velocity, or Force=Power/Velocity. A Watt is a Joule/sec or a Newton-meter/sec, so with 100% efficiency, the force exerted by the gear on the rack would be F=P/V = (15k N-m/sec)/1 m/sec =15kN. That's 15,000 Newtons, or roughly 3,370 pounds! Are you talking about a cog-wheel railway? Don't forget that the forces calculated above were for steady-state and 100% efficiency. The forces will be increased during acceleration/deceleration and due to inefficiencies, and you always need a safety factor, so a nice round value would be 5000 pounds.

Since you gave no formulas, it took me a while to figure out what the 60 meters was about (1m/s for 60 s), but the math is correct. To get a 12.7 cm PD gear to exert 5000 pounds of force on one or two teeth is going to require a gear made of unobtanium!

I think that the chap wants to generate 15kW using an alternator driven off a 1:10 gearbox. This is an output of 20 HP.

Hi,

Thanks, But still i am bit unable to understand.

Power=Torque * Angular Velocity

15K=torque * 2*pi*2.5 (Here 2.5 revolution per second, Since gear has 150RPM)

Torque=954.93 NM.

As per the Mitcalc.com, Spur gear calculation, for 15K watt if I have a spur Pinion and gear with RPM 150,112.5 respectively. I see the Pinion PD to be 90 mm and gear to be 120 mm diameter respectively.

Number of teeth for the pinion is 15 and gear is 20.

Face Width is 95mm for Pinion and 92 mm for Gear.

So My question is if i have a Rack and a pinion is the calculation would be similar as above.

"Torque=954.93 NM." That is correct. Now Torque is Force * Lever arm, or Force=Torque/Lever arm. For a gear with a PD of 12.7 cm, the lever arm is the pitch radius, or 6.35 cm, or 0.0635 m. F=T/L=955 Nm/0.0635 m = 15,000 N, as in the previous post.

I have the same impression as Baffled; that you are hoping to use a cogwheel railway to generate electricity. you will need 60 m of track for each minute of output, plus starting and stopping distance. That 1:10 gearbox is going to waste 10-15% of your energy, so now we need over 17,000 N of force. What magic source of energy are you going to use to bring the car back up to the top?

Thanks for your explanation.

I have a doubt, let say if we have a pinion and gear.Let say if we need to transfer power (15K watt) from gear to the pinion.

Consider pinion is 150RPM and Gear is 132.5 RPM.

Now is it possible to have pinion to be less than 105mm diameter?

As per the mitcalc.com (http://mitcalc.com/en/download.htm, Spur gear - external) software, the pinion diameter is less than 105 mm. Here how does it is possible to acheive 102mm and face width of 95mm.

then why can't we have a rack and pinion to acheive the same.

To help you visualize, here is a drawing of the 15 tooth, 12.7 cm PD gear with the corresponding rack:

As you can see, only two teeth are in contact, so those two teeth must exert ALL of the force required to transmit power. If you go to an even smaller pinion, there will be a point where only a single tooth must exert all the force, and due to the shorter lever arm (smaller pitch radius), the force required will be even greater. Remember that the torque calculations determine only the tangential component of force. There will be considerable friction forces of tooth sliding on tooth, which must be vectorially added to the tangential force, increasing the total force required.

Unfortunately, I don't have Windows installed on my Mac right now, so I can't currently verify your use of the calculator. On the other hand, any time you force a device to operate near its theoretical limits, you are asking for trouble. I personally would not design a machine where any component was knowingly required to operate above 50% of its theoretical limit.

At the scale you are describing, I presume the system is outdoors, where at a minimum, dirt is going to get into the teeth. The teeth ARE greased, aren't they? What happens if a pebble gets stuck in a tooth somewhere? If you would tell us what exactly you are trying to accomplish, then perhaps we could be more helpful...

Hi dkwarner,

Thanks for your great support, Actually i have taken a 50% efficiency only, i need to generate only 7.5Kwatts, I am using an alternator of 7.5KWatts only.

I am new to this feild, as a basic, I am just trying to understand to convert linear force to rotational force and then generate electricity.

Let say if i make the rack as like a pulley system, like rack surrounding on the gear, there will be more teeth in contact. will that help in transferring the power?

If you get chance to look into the mitcalc software (it is a excel based system) it would be great help for me.

Thanks in Advance,

regards,

Manjuanthan.S

By definition, a rack is linear. Since it is linear, it has ends, which means it must reciprocate (move back and forth). This in turn means it will produce pulses of power separated by no output at each end of motion.

If you bend a rack into a circle, you now have a gear with either internal or external teeth, depending on which way you bend it. The resulting gear will have many teeth, and must rotate very slowly, but since it has no end, you can now get continuous power from it, instead of pulses. It would be a lot of bending of the rack before it makes any appreciable change in tooth contact. Now the question is: what source of energy is going to provide the slow rotation?

What linear force do you have in mind.?

When I looked at mitcalc, the downloadable gear calculator was a .exe file, not a .xls file. If you can send me a copy of a .xls file, I can try it without having to install Windows. In fact I do mean to install Windows, hopefully in the next few days, but no promises...

Dick

thanks Once again, the excel sheet require's their dll to work properly, and i do not know whether we can share the excel sheet generated by mitcalc.com, since i haven't purchased their software. Right now i have it in demo mode.

I have one more question,

Let say i have made a (Belt configuration)mesh type gear,how does it will help to reduce the gear diameter.

Regards,

Manjunathan.S

With all spur gears, either internal or external teeth, the angular velocities are inversely proportional to the number of teeth, and the torques are directly proportional to the number of teeth (because the lever arm, which is the pitch radius, is directly proportional to the number of teeth).

Let's say you have your 12.7cm 15 tooth gear rotating at 150RPM and exerting 15kN of force. With the 0.0635m Pitch Radius, it will be exerting a torque of 955N-m. If the connecting gear has 150 teeth (10X as many), then that gear will have a pitch radius of 0.635m, rotate 15 RPM (1/10 as fast), feel the same 15kN of force, and have a torque of 9550N-m.

If the connecting gear has 1500 teeth (100X as many), then that gear will have a pitch radius of 6.35m, rotate 1.5 RPM (1/100 as fast), feel the same 15kN of force, and have a torque of 95,500N-m.

As you already pointed out, a smaller gear will rotate faster. As you can see below, when the gear has very few teeth, the bases of those teeth get quite narrow, so the teeth are weaker. You are also forced to use a smaller diameter shaft.

One last time: tell us what source of energy you have available, and we will be more helpful in figuring out what to do with it.

thanks dwarner,

This design is for the third party, They have a source mixed of disel and some renewamble energy(generated through waste). Most of the energy is used for their equipment. Some time the system is idle, there was a suggestion to use that for generating the electricity.

i enquired with them, linear motion is back and forth.

I would like to attachethe excel sheet which is generated from the mitcalc.com software, how could i attache the same?