Confused about Gear size and tooth number in relation to torque

I’ve been trying to do some physics calculations with vex, and am kinda confused. The torque equation is t=F*r, with f being the perpindicular force on the lever arm, and r being the distance from the pivot point. And from the vex website i found the gear diameters for a 12 tooth gear and a 60 tooth gear to be 1.5 cm and 6.5 cm respectively.
Knowing this, i would expect that in a 12:60 gear ratio, the angular velocity of the 60 tooth axle to be 1/5 as fast as the 12 tooth axle. And also that the 60 tooth axle would have 5 times as much torque as the 12 tooth axle. The vex website gives the stall torque of a vex motor as .734 newton meters.
However, when i calculated the torque with the given gear diameters, i found that the ratio between the 2 axles was .734Nm on the 12 tooth axle to 3.181 Nm on the 60 tooth axle. I then noticed that there was a discrepancy between the torque to torque ratio and the angular velocity to angular velocity ratio.
I still would expect the gear ratio to cause the 60 tooth gear to be 1/5 as fast, but i am confused on how the torque conversion isn’t inversly 5 times as well.

Can anyone help explain this?

my calculations for the torque are
T= Fr
T=.734 Nm
w(angular velocity) = 100rpm = 3.33 pie / seconds
.734 Nm= F (.0075m)

Force is constant on the next gear, so

T=97.87 N * .0325 m = 3.181 Nm

Mechanical Engineering is my weakest area, so I will trust your Equations…

Have you seen the Vex Robot Speed Chart??

Does any of this information correlate with your calculations???

Kevmaniac2000,

You are absolutely correct that the torque and speed ratios should be inverses of each other. If they weren’t, a gear train could be used to violate conservation of energy. (I’ll let you have the fun of deriving the proof of that.)

The difficulty you’re having seems to come from the definition of “diameter”. In practice, a gear has more than one “diameter”, and which one you use in calculations depends on what you’re trying to calculate.

The diameters you reported appear to be the outer diameters, that is, the diameters of circles that pass through the most distant points of the teeth. (I measured the outer diameters as 14.7 and 65.3 mm for those two gear sizes.) The outer diameter is important for calculating clearances, but it’s not the correct diameter for calculating either velocity or torque ratios.

For calculating those ratios, you need to consider the diameters at which the teeth of the two gears contact each other. If the gear train is working properly, that’s the “pitch diameter”.

For an excellent tutorial on gears, I refer you to:
http://www.cs.cmu.edu/~rapidproto/mechanisms/chpt7.html#HDR115
For a convenient table of gear formulae, see:
http://www.engineersedge.com/gear_formula.htm

The calculated pitch diameters of the 12- and 60-tooth gears, in mm, are 12.7 and 63.5, respectively.

The ratio of those two is, not coincidentally, 1:5. It’s not a coincidence, because pitch diameter is calculated from diametral pitch (24 for Vex gears) and number of teeth.

If you use half the pitch diameter as the lever length in your torque calculations (instead of half the outer diameter, which is what you appear to have done), you should get a torque ratio that is the inverse of the speed ratio.

Another approach is to consider the distance between gear centers. In the Vex system, the holes are on 12.7 mm spacing. That’s consistent with the sizes of most of the gears (The Vexplorer has a 48-tooth gear.), which have pitch diameters that are integral multiples of that. (BTW, if you do all the calculations in inches, the numbers are much “prettier”.) Once you fix the distance between centers, you can calculate the lengths of the lever arms by apportioning the space between centers based on the relative number of teeth. For the 12:60 example, divide the distance between centers (3 holes = 38.1 mm) by the total number of teeth (12 + 60 = 72) to get the “lever arm/tooth” of 0.529167. If you multiply that value by the number of teeth for each gear, you get 6.35 and 31.75, respectively. These are halves of the pitch diameter. That confirms our previous results.

If I can be of further help, please let me know.

Eric

It appears that you are using the outside (max) diameters of the gears in your calculations. If you use the diameter of the gear wheels halfway down the teeth you should find that your 5:1 assumption of Torque is verified. Gears are always modeled/calculated as circles at the point of contact of the teeth which is close enough to halfway down the teeth is reality and well designed gears.

so…
60 tooth wheel OD= 65.6mm ID=60.82mm (measured using Solidworks) so use diam 63.21. Rad=31.6mm

12 tooth wheel OD=14.8mm ID=10mm (measured using Solidworks) so use diam 12.4mm. Rad= 6.2mm

The ratio of the radii is 31.6:6.2 or 5.09:1. :smiley:

In reality this is 5:1

Your physics is correct in that the Torque is inversly proportional to the gear ratio.:smiley:

Cool!! I learned something today…

I learned something else today…

Thanks for your post… I definitely need to learn some more in this area…

Thanks! that’s really good to know, I’ll keep this in mind when i do more vex calculations later.:smiley:

you may also want to take a look at this page. it has all of the output torques for compound gear ratios. when i say preferred torque i mean the most efficient output for the motor (half of it’s stall torque) as max torque is the point the motor will stall.

by finding the needed torque of the motor to lift you arm or bucket or whatever you can then look at the chart to find the most efficient gear ratio for your needs. they are arranged in order of output torque.:wink:

https://vexforum.com/gallery/showimage.php?i=3236&catid=member&imageuser=8156

it seems like the preferred torque is half the stall torque, anyone know any reason why this is so?

Kevmaniac2000,
I believe the “preferred torque” is based on maximizing the mechanical power or efficiency (mechanical power out/electrical power in) of a motor. The mechanical power is the product of the rotational speed and the torque; the electrical power is the product of the current and voltage. At stall, you have the least possible efficiency, as the mechanical power is 0, and the electrical power is maximized. Similarly, the no-load speed produces no mechanical power (I don’t count the power to move the motor parts, as that can’t be extracted to do useful work.), because the available torque is zero. Therefore, the most efficient working point must be between them.

There are several good discussions of DC motor theory available on-line. I’d scare up a couple of links for you, but I suspect that would delay your seeing this message.

{I posted a rather detailed reply to your original question early Saturday, before any of the other replies. It is being held for moderator approval and I suspect that’s due to my embedding non-Vex links. If this message gets posted immediately, I may try re-sending the one from yesterday without the links.}

Eric

The best laid plans of mice and men…

I don’t seem to be able to find my way back to the long post with embedded external links, so I’ll type a short version of the easier ways to deal with torque ratios.

  1. If the gears are of the same pitch, then you can calculate both speed and torque ratios from the ratio of numbers of teeth. Frankly, as long as the gears are of the same pitch, I don’t see why you’d use any other method to calculate speed and torque ratios.

  2. If you want to know the diameter at which the gears are in contact, known as the “pitch diameter”, the easiest way to calculate it is to take the number of teeth and divide by the “diametric pitch” or “DP”. The Vex gears are all of DP = 24 teeth/inch. So, for example, the pitch diameter of a 12-tooth Vex gear is:
    (12 teeth)/(24 teeth/inch) = 0.5 inches.
    Similarly, the pitch diameter of the 60 tooth gear is:
    (60 teeth)/(24 teeth/inch) = 2.5 inches.
    The ratio of the pitch diameters is the same as the ratio of number of teeth, 5:1. This is not coincidence, it’s an algebraic result:
    PD2/PD1 = [T2/DP]/[T1/DP] = T2/T1
    for example:
    PD12/PD60 =
    (12 teeth)/(24 teeth/inch)]/(60 teeth)/(24 teeth/inch)]=
    (12 teeth)]/(60 teeth)]= 1/5

  3. Another way to calculate the effective radii is to start by considering the distance between centers of the two gears. In the Vex system, that is (assuming you have the gears aligned on a principal axis) an integral multiple of 0.5". The multiple is, as can be calculated using the formula above, the total number of teeth divided by 48 (twice the DP). The distance of the contact point from each center of rotation can be calculated by taking the center-to-center distance and distributing it to the gears in proportion to their numbers of teeth. Continuing our example:
    Center-to-center distance for 12/60 set =
    (12 + 60)/48 = 72/48 = 1.5 inches
    Dividing the distance by the sum of the smallest whole number elements of the gear ratio, in this case 1 + 5 = 6, gets us 0.25 inches. Therefore the contact radius of the 12-tooth gear is 0.25 inches and, of the 60-tooth gear, 1.25 inches. As you may well have noticed, these values are half of the respective pitch diameters.

Eric

That is to prevent Forum Spamming, since most Forum Spammers, create accounts and SPAM immediately.

If you get enough on topic Posts, this restriction is lifted…

You can always send a Private Message to the User, and Post the Forum Message Later, or send a Private Message to one of the Members or Senior Members with the Links, to re-post for you…

another cool way of doing this is to made a load speed curve of the motor. you can find examples online but basically you put speed on the horizontal axis (0-100 RMP will do) and then the load on the vertical axis (0-6.5 in/lbs) then mark a point at 0,6.5 and another at 100,0 and make a line between the two (a straight line is pretty much dead on) then by knowing the speed or the torque the axle is spinning at you can find the other by looking at the graph. it is because of this that you can find the power output. P out = rotational speed in radians per sec * Torque. i did a lab report on this so i will add on the section about this. i apologize if the tense is off i am just copy and pasting it.

lab report section:
ANALYSIS
LOAD. (See Load vs. Speed Graph) As the motor is loaded the angular speed of the motor decreases. The two extremes are the unloaded state where there is no load but maximum speed and stall torque were load is at its maximum but speed is 0 RPM. The relationship between load and speed is linear so if the motor is loaded to ½ its stall torque the angular speed will be ½ that if the unloaded speed. When the motor is loaded the electromagnetic interactions within the motor are being counteracted by the gravitational force on the load. This causes the motor to spin slower than if the electromagnetic interactions were not counter acted upon.
MECHANICAL POWER OUTPUT (See Mechanical Power Output vs. Speed and Mechanical Power Output vs. Torque) Mechanical power output does not remain constant as the motor is loaded. Mechanical Power Output = Torque * Speed (rad. per sec) on the two extremes you have stall torque were torque is high yet speed is zero so mechanical power is equal to zero. Here the electromagnet is not interacting with the stationary magnets and therefore produces no power. At unloaded speed the speed is high yet torque equals zero making mechanical power output equal to zero. Here the electromagnet passes through the field of the stationary magnets too quickly to effectively transform the electrical input into output power. At roughly half of the rotational speed range the electromagnet reaches a highly efficient state of interaction with the stationary magnets. Resulting in maximum output power. For my motor the maximum output power is equal to 78 mW at 420 RMP.
EFFIENCY The efficiency of my motor is not constant. My motor is most efficient when running at about 420 RPM (about ½ of its no load speed). Here the electromagnetic interaction with the stationary magnets is coupling most effectively. At the extremes the effectiveness of the motor (transducer) is zero. Although there is an electrical power input there is no mechanical power output. At peak efficiency my motor has an electrical input power of 2.22 Watts and a mechanical output power of about 78 mW (0.078 Watts) Which is an efficiency of 3.51% (sad but true). One idea I had to improve the efficiency would be to add more stationary magnets to more completely surround the electromagnet. This would increase the magnetic flux by adding additional magnetic field lines. This could also be accomplished by using stronger permanent magnets. This would increase the magnetic flux density. By using commercial bearings I could reduce mechanical friction within my motor increasing total efficiency. By using a magnet wire with higher conductivity I could reduce resistance allowing more current to do work and not be wasted in the form of heat.

Edit: i should prob. mention that for the lab i made a dc motor and tested it.