Why do triple bonds have a low NMR value?

Double bonds has a positive shift compared to single bonds. It seems like that there would be a trend but then the triple bond negatively shifts, despite having great electron density similar to the double bond. Any insight.

Chemical shifts arise due to differences in the local magnetic field in the different environments of a molecule. There are a number of contributing factors including the local diamagnetic contribution which is driven by the shielding effect of local electrons. Where there is functionality which generates electron circulation, a localised anisotropic magnetic field is created, and these can cause both downfield and upfield shifts, based on the position of the proton (nucleus) with respect to the functional group. These interactions, unsurprisingly, are called magnetic anisotropic effects.

The best example of this is the ring current effect arising from aromatic rings, with protons on the outside of an aromatic ring experiencing large downfield shifts, while protons either above the ring, or on the inside of the ring experiencing upfield shifts.

With double and triple bonds, the shielding cones are different in shape, and traditionally are drawn as shown below (you might find some better diagrams with a web search). Areas marked (+) increase the shielding (leading to decreased chemical shift) and areas marked (-) decrease the shielding (increasing chemical shift).

However, it is important to recognise that chemical shifts are not only influenced predominantly by electronics; sterics have a very significant contribution to chemical shifts also, often more so than electronics. A very good (recent) paper (Baranac-Stojanovic, M RSC Adv., 2014,4, 308-321) discusses traditional magnetic anisotropy models, and presents some very good examples that highlight the effects of steric influence on chemical shifts.

Chemical shift, $\delta$, is defined by $${\delta\equiv{(\nu_S - \nu_R)\over\nu_R}\times10^6}$$ where $\nu_S$, and $\nu_R$ are the resonance frequencies from sample and reference, respectively. Both $\nu_S$, and $\nu_R$ follow Larmor equation $${\nu_0={\gamma\over2\pi}B_0(1-\sigma)}$$ where $\gamma$ is the gyromagnetic ratio (or is called magnetogyric ratio), which is nucleus specific, and $B_0$ is the magnetic field strength. $\sigma$ is shielding factor, or chemical shielding. For a given nucleus, such as $\ce{^1H}$ or $\ce{^13C}$ in a certain magnetic field $B_0$, for example $9.4T$ ($\ce{^1H}$ resonates at about 400 MHz), the chemical shift difference depends only on the shielding factor $\sigma$.

Chemical shift can be rewritten as $$\delta={\sigma_R-\sigma_S\over 1-\sigma_R} \times 10^6\approx (\sigma_R-\sigma_S)\times10^6$$ because $\sigma << 1$

The change in shielding factor can be caused by variation in electron density. When observed nucleus is near a strongly electronegative group, the shielding factor $\sigma_S$ is smaller (less shielding), chemical shift is larger. Based on this alone, triple bonds would be less shielded and have larger chemical shifts than double bonds. But unfortunately, electron density is not the only factor that affect the chemical shielding.

Anisotropy effect is due to the anisotropy in the magnetic susceptibility of nearby groups of the observed nucleus. Because of this effect, triple bonds have a much larger chemical shielding than double bonds along the direction parallel to $B_0$. There are equations describing change in shielding and magnetic susceptibility tensors. But I don't think I am capable to explain any deeper.

Table. Some Calculated and Experimental $\ce{^13C}$ Shieldings in parts per million $$\begin{array}{c|c|c|c|c|c|} & \ce{\sigma_{xx}}&{\sigma_{yy}} & \ce{\sigma_{zz}} & \ce{\sigma_{av}}&\text{Expt.} \\ \hline \ce{CH4} & 195.8 & 195.8 & 195.8 & 195.8 & 195.1 \\ \hline \ce{C2H6} & 187.7 & 182.7 & 193.1 & 186.2 & 180.9 \\ \hline \ce{C2H4} & 177.9 & -81.1 & 84.3 & 60.4 & 64.5 \\ \hline \ce{C2H2} & 39.0 & 39.0 & 279.4 & 119.1 & 117.2 \\ \hline \end{array}$$

The shielding $\sigma_{av}$ corresponds to $\sigma_S$ in the above chemical shift equation. $\sigma_{zz}$ has a direction parallel to $B_0$.

Reference "High Resolution NMR: Theory and Chemical Applications" (3rd Ed) by Edwin D. Becker

(The table is from the above reference too.)