1. ## Position tolerance exercises

Hello,

I'm solving some question about position tolerance from a book and there are two of them which I didn't understand what is wrong with my resolution. Could someone help me with it?

Picture for question 47 and 48:

Capture.PNG

47. For the feature control frame marked A, how much datum feature shift is allowed if the datum feature is produced at 14.8 and has 0.5 perpendicularity error?
A. 0
C. 0.5
D. 0.9
E. 1
F. 1.2

My resolution:

Datum feature shift= MMB - rAME

MMB= MMC + Geo tol. (respecting datum precedence) = 15.2 - 0 (No geo. tol. is added since datum B (MMC) is primary datum and perpendicularity tolerance applied to B depends on A) = 15.2

rAME = uAME + perpendicularity error = 14.8 + 0.5 = 15.3

Datum feature shift= 15.2 - 15.3 = -0.1 (This value tells me that there is a inteference of 0.1 which is the opposite of datum shift porpuse)

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48. For the feature control frame marked A, how much datum feature shift is allowed if the datum feature is produced at 15.2 and has zero perpendicularity error?
A. 0
B. 0.4
C. 0.5 (official awnser)
D. 0.9
E. 1
F. 1.2

My resolution:

Datum feature shift= MMB - rAME

MMB= MMC + Geo tol. (respecting datum precedence) = 15.2 - 0 (No geo. tol. is added since datum B (MMC) is primary datum and perpendicularity tolerance applied to B depends on A) = 15.2

rAME = uAME + perpendicularity error = 15.2 + 0.5 = 15.2

Datum feature shift= 15.2 - 15.2 = 0

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In fact, both question seems to add the perpendicularity tolerance to MMB but isn't it breaking the datum precedence?

PS: I'm studying GD&T by myself trough the standard and some books. So I don't have any teacher guiding me who could solve the exercises.  Reply With Quote

2. Who's book is this from?  Reply With Quote

datum precendece, position toleracing 