Snail Cam shaft torque calculations

[teun-lll]

TL;DR

I need help with finding the right motor parameters for my cam system

Hello,

Im trying to calculate the torque needed for a system that i want to design. i want to raise and drop a weight of 800 kg.
Ive made some calculation but I am not certain i did it right. I think I am on the right track but i feel like the numbers are a bit high. I have a hard time finding good examples to confirm my answer. Could someone take a quick look at it?
Thanks!

 

[BvU]

Hello,

If I understand your picture well, you want to lift the mass by 100 mm in 2$\pi$ radians ?
Then I don't see where the 300 mm comes from ?

Work done would be $\ mg\Delta h = 800$ J, which is $\ 800/ 2\pi$ J/radian.
In other words: the required torque is $127$ N.m

Lifting 800 kg by 0.1 m in 2 seconds requires $P = mgh/\Delta t = 400$ W

There are a few considerations:
1. the calculation is without acceleration (other than $g$) so inelasticity and high rpm are not included.
2. The proposed setup looks more like a design to quickly ruin the bearings of the cam than anything else !

[edit] but of course a bottom for the load to drop on to can overcome that. I liked your prototype mechanism!

$\$

 

[berkeman]

What is the application? What happens if the lifting apparatus fails and the load drops?

 

[BvU]

He wants to rattle and shake big bags to compress the contents !

 

[berkeman]

Ah, I wondered if this was related to his previous thread. Thanks.

 

[teun-lll]

yep it is related. good research. i want to lift the mass 150 mm. The 300 mm is the max distance. so the snail cam has 150 mm drop and from there another 150 mm to the center. as shown in the picture.

The design i showed is a very basic version of what i want to do. But in principle this is what i need. i have a solution to save the bearings ;)

@BvU i don't understand how you can use a simple equation like that when you don't look at the angle, and shape of the cam

could you also explain where my calculation failed. because it does seem logical to me.

Edit: exactly. i want springs to break the fall before it hits the cam. I also liked the prototype and i made a proposal for it. But since it uses quite a few custom parts, it would become too expensive.

 

[jrmichler]

If I understand correctly:
1) The maximum cam radius is 300 mm.
2) The minimum cam radius is 150 mm.
3) The cam surface is at an angle of 81 degrees to the radius at the 300 mm radius.
4) You want 30 RPM.

Then:
1) The static normal force is $8000 N / cos(9 degrees) = 8100 N$. This is the design load on the cam roller bearing.
2) The force component perpendicular to the radius is $8100 * sin(9 degrees) = 1270 N$.
3) That force is at a radius of 300 mm = 0.3 m.
4) The torque is $1270 N * 0.3 m = 380 Nm$.
5) Power = $380 Nm X 0.5 rev / sec * 2 * PI radians/revolution = 1200 Nm/sec = 1200 Watts$.
6) Safety factor of 2: $1200 X 2 = 2400 watts = 3.2 hp$.

Some suggestions:
1) Gear box efficiency needs to be part of the required power calculation. This is especially important for worm gear boxes.
2) Good practice is to mount the cam on a separate shaft supported by easily replaceable bearings, such as pillow block bearings. The shaft is connected to the gearbox output shaft by a shaft coupling.
3) Oversize your cam roller bearing. Then design in a spring mount so that it won't get destroyed if the falling load lands on the cam roller bearing.

 

[BvU]

@BvU i don't understand how you can use a simple equation like that when you don't look at the angle, and shape of the cam

I looked at where the energy went...
(but initially concluded that the X at 15 was for 0.15 m and therefore the lifting/revolution was from 0.1 to 0.2 -- so a factor of 1.5 is needed. Times 2 for the safety margin )

could you also explain where my calculation failed. because it does seem logical to me.

You forgot to set your calculator to degrees ...
So perhaps you want to explain how you managed to come up with the 7 degrees ?
(Which grew to 9 degrees in #7 )

And I become very suspicious if forces increase when split into components...
But perhaps @jrmichler can explain ...
$\$

 

[teun-lll]

Great, thanks guys! Yes somehow the numbers seem to change along the way. But i recon he means 7 degrees.
Thx for pointing out that my calculator wasnt set to degrees

7 degrees is what i measured when taking two points very close to each other and creating a line. i did this at multible points and got a steady 7 degrees, relative to the perpendicular line coming from the center to that point. So i thought it was sufficient enough to work with.

I now see where i went wrong(besides my calculator): i totally forgot this step:
2) The force component perpendicular to the radius is 8100∗sin(9degrees)=1270N.
I will redo my calculations with the given information.

"And I become very suspicious if forces increase when split into components...
But perhaps @jrmichler can explain ... "

I understand what you mean, but in this case the number has to increase because the angle makes it harder to move the mass vertically then when you apply the force straight in the direction of movement. I don't know how to explain better but the calculated Normal force is the force that can be split up in the 8000 N and the perpendicular force.

I enjoy this discussion. Thanks and i will update after i revised my calculations.

 

[BvU]

7 degrees is what i measured when taking two points very close to each other and creating a line. i did this at multible points and got a steady 7 degrees, relative to the perpendicular line coming from the center to that point. So i thought it was sufficient enough to work with.

In your pdf from #1, I now understand: R is growing from 1 = 150 mm to 2 = 300 mm in one revolution = $2\pi$ radians. In short:$$
R = R_0 \Bigl (1+{\phi\over 2\pi} \Bigr )$$ with $R_0 = 0.15$ m. $\quad$ Correct me if I am wrong.

Simple differentiation learns that $\displaystyle {{dR\over d\phi} = {R_0\over 2\pi}}$ .

And the graph $\ R(\phi)\$ of course looks trivial. It gave me the idea to replace the cam by a sawtooth, not for realization but to make it easier to understand (for me, that is ...). No more problem with $\ N > mg\$ !

$F = mg \tan\alpha\$ has to drag to the right (i.e. the wedge has to be pulled to the left) over a distance $\ R_0 /\tan\alpha \$ to lift the load by $\ R_0\$ . Work done is $\ mgR_0\$ (keeps me happy).

As @jrmichler indicates, associated torque at the end is $\ 2R_0F\$. But average is 25% less and I figure a flywheel can smear it out for the drive motor ?

My problem/confusion is to understand where the 7 degrees comes from. I see it in Teun's picture, I do it numerically in a messy XL sheet and find 7.2, 9.1, 7.9, 7.0, 6.4 degrees at 0, ½π, π, 1½π, 2π so I'm doing something wrong (?). It's even worse when I try to do it with symbols:

​

Go from $\ \phi\$ to $\ \phi+d\phi \$: tangentially $\ R\,d\phi\$ and radially (see above) $\ dR = R_0/(2\pi)\, d\phi\$ so a tangent of $${R_0\over 2\pi R} = {R_0\over 2\pi R_0(1+\phi/(2\pi)} = {1\over 2\pi+\phi }$$meaning an angle that decreases from 9 to 4.5 degrees. Would be nice for a constant torque but I don't believe it...​

​

Can I ping @haruspex to put me right here...?

$\$

 

[haruspex]

R is growing from 1 = 150 mm to 2 = 300 mm in one revolution = $2\pi$ radians. In short:$$
R = R_0 \Bigl (1+{\phi\over 2\pi} \Bigr )$$ with $R_0 = 0.15$ m. $\quad$

If the angle of the surface to the radius is constant then it is a geometric spiral, $r=r_0e^{k\theta}$, where $k=\frac 1{2\pi}\ln(2)$.
The "slope" is given by $\frac 1r\frac{dr}{d\theta}=k=6.32°$

 

[haruspex]

If you want constant torque, $\tau$, it will not be a constant angle. You want the same work done in equal increments of rotation, so it is an arithmetic spiral: $r(\theta)=r_0+\frac{\tau}{mg}\theta$.
You also have $r(2\pi)=2r_0$, so $\frac{\tau}{mg}2\pi=r_0$.

 

[teun-lll]

If you want constant torque, $\tau$, it will not be a constant angle. You want the same work done in equal increments of rotation, so it is an arithmetic spiral: $r(\theta)=r_0+\frac{\tau}{mg}\theta$.
You also have $r(2\pi)=2r_0$, so $\frac{\tau}{mg}2\pi=r_0$.

In my calculations your last formula gives 1849.14 N/m. this seems to be high.
I now realize that its logical that the angle can't be the same everywhere if you need constant torque.

 

[haruspex]

In my calculations your last formula gives 1849.14 N/m. this seems to be high.
I now realize that its logical that the angle can't be the same everywhere if you need constant torque.

Is that the torque? Torque should have units of Nm, not N/m.
How do you arrive at that number? τ=2πmgr₀=2π×800×9.8×0.15Nm.