help pignone

[nicolapastena]

guys can't set a tooth sprocket from the toolbox with the data I have:
rp=9.55mm (primitive gear) module 1.75 helical inclination 20°
please if someone knows how to help you would be able I have already seen around and on the help of solid works 2011 thanks

 

[meccanicamg]

guys can't set a tooth sprocket from the toolbox with the data I have:
rp=9.55mm (primitive gear) module 1.75 helical inclination 20°
please if someone knows how to help you would be able I have already seen around and on the help of solid works 2011 thanks

Obviously the input parameters are different:

- number of teeth
- form
- pressure angle
- helical angle

the rest lives alone.:finger:

 

[sampom]

guys can't set a tooth sprocket from the toolbox with the data I have:
rp=9.55mm (primitive gear) module 1.75 helical inclination 20°
please if someone knows how to help you would be able I have already seen around and on the help of solid works 2011 thanks

you can't because to choose from the toolbox you need to know the nr of teeth. and with that primitive radius making the calculation with the module do not get an entire number.
you could approximate to z11 (if you can fit) and so the selection from toolbox works.
or you have to build one with a modified toothing or with a module "out of standard"; Copi and incolli (save by name) a toolbox gear and change the sketch odds to your use.
however you must always start from a no. of known teeth (or revenues from the reverse calculation between øp and module) and clearly whole.

what relationship do you have to get and with what intersect?

greetings
Mar

 

[nicolapastena]

I must have a ratio of 12/10 between pinion and rack (I know that at a rotation of 130° of the pinion corresponds a shift of the rack of 25mm, attention these are the project data that I have to respect) so I hypothesized a step that makes me have a number of teeth equal to 12 for the pinion and 10 for the rack (pass=5mm).

 

[meccanicamg]

I must have a ratio of 12/10 between pinion and rack (I know that at a rotation of 130° of the pinion corresponds a shift of the rack of 25mm, attention these are the project data that I have to respect) so I hypothesized a step that makes me have a number of teeth equal to 12 for the pinion and 10 for the rack (pass=5mm).

Sorry but the pinion and rack does not have a defined transmission ratio. is not a multiplier used in that way. a lazy lap corresponds to the number of linearized teeth on the rack.

pinion couple = axial force rack * dp/2

with dp = primitive diameter

length cremagleira = n*zp*p = n*dp*3.14
with n = number of turns or fraction
with zp = number of lazy teeth
with p = step = 3.14* module

effective length = rack length + mechanical extractor
with recommended mechanical extracorsa = 1/2 lap

review the choices you made

 

[nicolapastena]

the project data are that at 130° rotation of the pinion must correspond a translation of the rack of 25mm.
the pinion will rotate 130° to the right and to the left for which the total length of the rack will be 50mm.
appearance tips

 

[nicolapastena]

I cannot introduce an extracurrency otherwise the system (steel box) would no longer work for the prescribed data but would have a greater steering.

 

[meccanicamg]

I cannot introduce an extracurrency otherwise the system (steel box) would no longer work for the prescribed data but would have a greater steering.

mechanically you can't get a pinion for a tooth on the rack, otherwise it breaks. extracorsa serves as a physical/mechanical margin to ensure n teeth intake. must not be done as a race.

if a spin of pinion hypotheses are 10 teeth, you will make 10 teeth forward and 10 teeth back on a 10+10+5 teeth long rack.

:finger:

the race of the rack is 25+25 but can be also 1 meter long the cremagleira

 

[nicolapastena]

is true in the end the end of the race is mechanical,, with regard to the reasoning of the hypothesized step and the number of teeth according to you all right?

 

[sampom]

...so that at a rotation of 130° of the pinion corresponds a shift of the rack of 25mm, attention these are the data of project that I must respect. . .

Then tell me where you pulled that one out. rp=9.55?
with that radius in 130° the rack would move only 21,668 mm.
to have 25mm you need a øp of 22,0368 and as I said before you will have to build "special" sprockets and racks, because the teeth are "physical elements" and clearly can only be in full number.
or if you agree to approximate with standard modules you can use a z11 m2 and get 24.958mm moving to the rack.

..but do you understand how gears, transmission reports and racks work?

greetings
Marco:smile:

 

[sampom]

..but do you understand how gears, transmission reports and racks work?

Sorry but the pinion and rack does not have a defined transmission ratio. is not a multiplier used in that way. a lazy lap corresponds to the number of linearized teeth on the rack.
review the choices you made

..right:

greetings
Marco:smile:

 

[nicolapastena]

excuse me then nn I remember the angle if it was 130°, it is safe cmq that the radius is 9,55mm. and to create a special pinion as I do with solidworks?

 

[meccanicamg]

quoto sampom....

parts from the reasoning of the stroke length = n turns * zp * lazy * mn

and fixed that:

- stroke length = 25 mm
- n r r r = 130°/360 [°/giro] = 0.3611111 approximated to 0.36 rpm

now you have to fix or the number of zp teeth or fix the normal module mn. see what to do and how to round the digits and especially how all sizes have a +/- or +/+ or -/- then manage the thing in a more complete way.

 

[nicolapastena]

I ask vein is 150° for 25 mm translation:
♪
t=translation of the rack
a=rotation angle
r=r
I get r=9.55mm

 

[sampom]

Excuse me then I remember the angle if it were 130°, it is safe cmq that the radius is 9.55mm.

if you are sure of that radius and the shift of 25mm then the angle will be 149.99°. but the fact remains that standard gears of that diameter do not exist because impossible (they are simple geometry calculations of the circle).

and to create a special pinion like I do with solidworks?

I told you a few posts above.

greetings
Mar

 

[meccanicamg]

excuse me then nn I remember the angle if it was 130°, it is safe cmq that the radius is 9,55mm. and to create a special pinion as I do with solidworks?

to change a toolbox bookbine just save it by name as a copy. open your saved file and change equations. There's gonna be a little bit to get out of here because you need to get some equations and fix everything. That's the way.

But it is not clear to me if you know what you have to do or if it is a random series of questions.

 

[sampom]

if you are sure of that radius and the shift of 25mm then the angle will be 149.99°.

I ask vein is 150° for 25 mm translation:

Here.
are not however 150° just but 149.99

greetings
Mar

 

[nicolapastena]

after what I did:
ipotizzo un passo p=5mm p=(2*3.14*r)/zp
lazy step = racket step
zp=num pinion teeth=12
zc=number of rack teeth=50/5=5

 

[sampom]

But it is not clear to me if you know what you have to do or if it is a random series of questions.

I'm afraid the second you said.....:frown:

 

[nicolapastena]

I'm not worried about you guys if I'm going to get out of here anyway. I'm sorry I didn't exactly do that.
for what I wrote, do we agree on the number of teeth I hypothesized?
moving to the table on solidworks tell me to approximate to one of a toolbox or make it ad hoc?