How to Linearly Combine Two Unbiased Estimators of One Parameter without Knowledge of Their CoVariance?
Basically I'd do something like:

$$ \frac{ {\sigma}_{2}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2}} {r}_{1} + \frac{ {\sigma}_{1}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2}} {r}_{2} $$

This is the optimal weighing given knowledge of the Variance only (Well, linear).
Basically it assumes the cross correlation is 0.

Derivation

There are many ways to derive this.
The easiest would be intuition of knowing that Variance is the Linear Measure of the quality of an estimator. But we're leaving intuition our for this moment.

So easiest formal derivation could be using the ideas @Mark Leeds talked about.

Our model is:

$$ {r}_{3} = {w}_{1} {r}_{1} + {w}_{2} {r}_{2} $$

We want to guarantee the mean of $ {r}_{3} $ is un biased so:

$$ \mathbb{E} \left[ {r}_{3} \right] = \mathbb{E} \left[ {r}_{1} \right] = \mathbb{E} \left[ {r}_{2} \right] = {w}_{1} + \mathbb{E} \left[ {r}_{1} \right] + {w}_{2} + \mathbb{E} \left[ {r}_{2} \right] \Rightarrow {w}_{1} + {w}_{2} = 1 $$

So we'll use $ {w}_{2} = 1 - {w}_{1} $. Now we want to minimize its variance. Since it is not biased we're after minimizing its second moment:

$$\begin{align*}
\mathbb{E} \left[ {r}_{3}^{2} \right] &amp; = \mathbb{E} \left[ { \left( {w}_{1} {r}_{1} + \left( 1 - {w}_{1} \right) {r}_{2} \right) }^{2} \right] \\
&amp; = {w}_{1}^{2} \mathbb{E} \left[ {r}_{1}^{2} \right] + { \left( 1 - {w}_{1} \right) }^{2} \mathbb{E} \left[ {r}_{2}^{2} \right] + 2 {w}_{1} { \left( 1 - {w}_{1} \right) } \mathbb{E} \left[ {r}_{1} {r}_{2} \right] \\
&amp; = {w}_{1}^{2} {\sigma}_{1}^{2} + {\left( 1- {w}_{1} \right)}^{2} {\sigma}_{2}^{2}
\end{align*}$$ 

We assumed $ \mathbb{E} \left[ {r}_{1} {r}_{2} \right] = 0 $.

Now, this is a quadratic form so to minimize just derive it and find what vanishes it:

$$ {w}_{1} {\sigma}_{1}^{2} - {\sigma}_{2}^{2} + {w}_{1} {\sigma}_{2}^{2} = 0 \Rightarrow {w}_{1} = \frac{ {\sigma}_{2}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2} }, \; {w}_{2} = \frac{ {\sigma}_{1}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2} } $$

Pay attention that the solution doesn't require us to know the exact variances of the estimators being fused. Only their ratio.

So in the case above where $ {\sigma}_{1}^{2} : {\sigma}_{2}^{2} = 0.8 : 0.2 $ the solution is $ {w}_{1} = 0.2 $ and $ {w}_{2} = 0.8 $. This is the best you can do without the knowledge of the co variance.

This fused veritable will have variance $ {\sigma}_{3}^{2} = {0.2}^{2} {\sigma}_{1}^{2} + {0.8}^{2} {\sigma}_{2}^{2} = 0.16 \hat{r} $

Wikipedia Solution

You may, as a reference, look at Wikipedia - Sensor Fusion.
They wrote the same equation in a real clumsy manner:



Yet they are equivalent (Just multiply their result by $ \frac{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2}} $)

Pay attention that the sensor model assumes that 2 sensors are not correlated (Well, it assumes noise is property of the measurement equipment, hence this assumption is made).