Question

The Distance between end projection of the line AB is 70 mm and the projection through traces are 110 m apart. the end of the line 10 mm above HP. If the top view and front view of the line makes 30°and 60°with the line respectively. draw the projections of the the l line and determine the traces, the angles of Hp and VP and the true length of line​

Answer 1

Answer:

$\begin{tabular}{||l} \mathrm{TL}=146 \mathrm{~mm} ; \theta=56^0 ; \Phi=16^0 ; \alpha=60^0 ; \beta=30^0 \\ \mathrm{VT}=192 above \mathrm{HP} ; \mathrm{HT}=64 in front of \mathrm{VP} .\end{tabular}$

Explanation:

Step 1: Dist between the projectors $\left(\mathbf{d}_{\mathrm{p}}\right) \quad=70$

Dist between the $\underline{\text { traces }}\left(\mathbf{d}_{\mathrm{T}}\right) \quad=\mathbf{1 1 0}$

End $\mathbf{A}$ from HP (a') (above HP) $=\mathbf{1 0}$

$\boldsymbol{\alpha}$ (angle made by FV with xy) $\quad=\mathbf{6 0}^{\circ}$

$\boldsymbol{\beta}$ (angle made by TV with xy) $\quad=\mathbf{3 0}^{\circ}$

Step 2: Whenever distance between traces $\left(\mathrm{d}_{\mathrm{T}}\right)$ is given, draw two vertical lines at the given distance to mark$\mathrm{h} \& \mathrm{v}$ on $\mathrm{x}-\mathrm{y}$ where these two lines cut $\mathrm{x}-\mathrm{y}$.

Join $(h, V T) \&(v, H T)$. On these lines, the Front View (FV) \& Top View (TV) will lie.

Locate starting point a' \& then draw 2 vertical lines $\left(\mathrm{d}_{\mathrm{p}}\right)$ starting from a' which will cut $(h, V T) \&(v, H T)$.

Since the FV \& TV lie between $d_p$, we can get the FV \& TV. Then the True Length (TL) \& angles can be found.

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