Modulo meccanico by palaia

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· @palaia ·
$3.47
Modulo meccanico
https://youtu.be/cr_eJJypCPM


<div class="text-justify"> 
<center>

Apriamo una vista dall'alto dove andremo a dare una prima quota di diametro 25mm (d.a.) si aggiungerà un (d.)e un (d.f.) misure di diametri occorrenti per il calcolo della ruota dentata.


<center>![Screenshot_2018-03-14-21-57-38-1.png](https://steemitimages.com/DQmWPNK9bYjcXvx5iLhShY2QnDXHENuR1acQngLpytYgBNQ/Screenshot_2018-03-14-21-57-38-1.png) </center>


Per un pefetto funzionamento si necessitano delle formule esatte, senza di esse non si avrà un movimento rotatoio conguaglio.


<center>![Screenshot_2018-03-14-21-55-23-1.png](https://steemitimages.com/DQmdMEC6qHHujgH5FARz6vTbNrLPMSBHEeAK96k5JfXCRth/Screenshot_2018-03-14-21-55-23-1.png) </center>



Ha=Altezza dente esterna
Hf=Altezza dente interna
H=Altezza dente totale
C=Tolleranza tra le parti.
Formula da mettere in atto :

Z=D÷M
Ha=M
Hf=M+C
H=2×M+C
C=Nella maggior parte dei casi e quasi sempre 0.167×M
Questa è una prima delle formule il resto sarà nel passi successivi.


<center>![Screenshot_2018-03-14-21-56-55-1.png](https://steemitimages.com/DQmbd1apyxgN2MDCG8ENBSYyfLrTQeJHiJWd1mAAfsXHVkA/Screenshot_2018-03-14-21-56-55-1.png)</center>


Tra i due assiemi di motrice e condotta (bullone) si necissita una distanza chiamata (c) che sarebbe una tolleranza tra le due parti (estremità).


<center>![Screenshot_2018-03-14-21-55-28-1.png](https://steemitimages.com/DQmSmPf5XUUGKzX8LtQpJHakvCXCUzmFtimUWCDVVjD1Drn/Screenshot_2018-03-14-21-55-28-1.png) </center>


Vediamo come realizzare il bullone vedi foto:


<center>![Screenshot_2018-03-14-21-56-00-1.png](https://steemitimages.com/DQmP5xnNSBjYpRfc5Fs4ek54ka9rGzTJf8fbkFTrZ6Fu9iG/Screenshot_2018-03-14-21-56-00-1.png)</center>



<center>![Screenshot_2018-03-14-21-56-13-1.png](https://steemitimages.com/DQmSQptTKwaLAyZBFYknLxS8ohhJ7fYEDNoEojipd4m2p4H/Screenshot_2018-03-14-21-56-13-1.png)</center>



Come prima cosa si dovrà sapere il diametro per farlo combaciare con la ruota dentata nel nostro caso il bullone ha un diametro di 7mm.


<center>![Screenshot_2018-03-14-21-58-06-1.png](https://steemitimages.com/DQmTX1jVcU6Hf3Earuri1Y7vouFGfJfUdT1Ltmvjw5Nj1mE/Screenshot_2018-03-14-21-58-06-1.png)</center>



Da una vista dall'alto creiamo il il primo cerchio.



<center>![Screenshot_2018-03-14-21-58-31-1.png](https://steemitimages.com/DQmPdTSwMK8vJqjVdsUPViZS2dMVGigubLggk6nm7Q4cqG7/Screenshot_2018-03-14-21-58-31-1.png)</center>


Mentre il secondo cerchio di 6mm sara il primo l'incastro dove andremo a fissarlo nella scatola per dare una posizione di movimento nelle entrambi direzioni avanti e indietro.



<center>![Screenshot_2018-03-14-21-58-11-1.png](https://steemitimages.com/DQmVs2iqSgeAfmEXBMKwmfZnX929bgubcCfMQTD8upC6Zwa/Screenshot_2018-03-14-21-58-11-1.png)</center>



Una volta dato la giusta profondità di necessità il secondo cerchio di diametro 6mm per un incastro completo ,come si può vedere nelle due foto in basso.



<center>![Screenshot_2018-03-14-21-55-18-1.png](https://steemitimages.com/DQmcBVbrfGKvKRUwi9MjgVp1EjxzgQNhpf6SN7BwdPRp29y/Screenshot_2018-03-14-21-55-18-1.png) </center>


<center>![Screenshot_2018-03-14-21-58-11-1.png](https://steemitimages.com/DQmVs2iqSgeAfmEXBMKwmfZnX929bgubcCfMQTD8upC6Zwa/Screenshot_2018-03-14-21-58-11-1.png) </center>


Una volta creato la sua base liscia bisognia dare una quota a spirale per far si che diventi un bullone vero e propio si necessita una linea di mezzeria più un cerchio dello stesso diametro ,in questo caso diametro 7mm.


<center>![Screenshot_2018-03-14-21-57-43-1.png](https://steemitimages.com/DQmVPCUsvWHoiDUQXmvM3wKiKJR7os7t2UkSz2iw75kB2e6/Screenshot_2018-03-14-21-57-43-1.png) </center>



Dopo aver fatto la spirale bisognia creare una incavatura per far si che abbia la sua funzione come bullone.
Queste misure sono di mio piacimento non ho usato le formule di riferimento per il semplice fatto che ne ho voluto inventare una io.
Capita spesso che alcune società inventino nuovi moduli per una questione di vendita sul mercato e per non essere concorrenti con le altre .


<center>![Screenshot_2018-03-14-21-57-49-1.png](https://steemitimages.com/DQmPZpNCFo3nijzQRGLK5V311HdNS5Qphrx1TdDkK2be1mC/Screenshot_2018-03-14-21-57-49-1.png) </center>



Dopo di che si potrà attivare l'avvolgimento a spirale dove avremo questo risultato :


<center>![Screenshot_2018-03-14-21-58-19-1.png](https://steemitimages.com/DQmasUzyRQD17JrS3GHUGCKMKr9sz2eBCV1AfKyWec62btr/Screenshot_2018-03-14-21-58-19-1.png)</center>


<center>![Screenshot_2018-03-14-21-57-31-1.png](https://steemitimages.com/DQmQ8CjwJhn1HYwmDg271ZgATqVUVxoKjwkBbSTBaLP74Vo/Screenshot_2018-03-14-21-57-31-1.png) </center>


Una volta creato i due primi assiemi più importanti, si potrà passare alle basi dove inserire entrambi componenti per il suo funzionamento.

<center>![Screenshot_2018-03-14-21-55-45-1.png](https://steemitimages.com/DQmVZCFfoFxkDLgJThjfdJ7Rm6CEimnEJE5omahcBgz5Dxm/Screenshot_2018-03-14-21-55-45-1.png) </center>


Partiamo da una base


<center>![Screenshot_2018-03-14-21-55-51-1.png](https://steemitimages.com/DQmcBzK1kHwE2JAwKgcuFEmcuU771YAE5xW7E8FgPx3wf6y/Screenshot_2018-03-14-21-55-51-1.png)</center>


<center>![Screenshot_2018-03-14-21-57-00-1.png](https://steemitimages.com/DQmTT2HEDzTm8QnuAdue54jaefZY6CQTmwuahFepN2R5AQm/Screenshot_2018-03-14-21-57-00-1.png)</center>


Ottenendo questo risultato.


<center>![Screenshot_2018-03-14-21-55-35-1.png](https://steemitimages.com/DQmWNsJxH8YaJENAKr8symCyLHHkJGjw3NpcNa5FLoLjAQi/Screenshot_2018-03-14-21-55-35-1.png)</center>


<center>![Screenshot_2018-03-14-21-55-28-1.png](https://steemitimages.com/DQmSmPf5XUUGKzX8LtQpJHakvCXCUzmFtimUWCDVVjD1Drn/Screenshot_2018-03-14-21-55-28-1.png) </center>


M=Modulo
P=(Teilung)distanza di passo degli ingranaggi.
D=(Teilkreis)  diametro detto anche diametro primitivo.
Da=(Kopfkreis) diametro esterno
Df=(Fuskreis) diametro interno

Le sue formule sono:
M=D:Z 
P=3,14×M
D=M×Z oppure Z×P÷3.14.
Da=D+2×M oppure M×(Z+2) 
Df=D-2×(M+C) 


Sotto si può vedere l 'assieme di posizione dove verrà montato il tutto sulla sua superfice :


<center>![Screenshot_2018-03-14-21-55-55-1.png](https://steemitimages.com/DQmQiEAJyiwDYM45Rx5kUyW1yRxc7B1qHspPKJDCwBmunR2/Screenshot_2018-03-14-21-55-55-1.png) </center>


Tra ruota dentata e bullone la loro distanza viene nominata con la lettera (a) sarebbe la metà di entrambi i pezzi dove la sua formula e:
a=d2-d1÷2=M×(Z2-Z1)÷2
D2 corrisponde alla ruota dentata 
D1 al bullone.

</div>

<center>Foto disegnio video di mia proprietà </center>
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