What's the "typical" tension necessary to keep a 88ft doublet aloft?
I'll also be supporting the midpoint where the feedline attaches.
Insulated 18awg will be used as the antenna wire.

tnx
jtm

On 3 Mar 2005 06:54:30 -0800, "Jim Miller" wrote:
What's the "typical" tension necessary to keep a 88ft doublet aloft?

Hi Jim,

It is the weight of the wire times the arctan of the angle the sag of
the wire depresses from the horizontal if I recall correctly.

73's
Richard Clark, KB7QHC

Jim Miller wrote:
"What`s the "typical" tension necessary to keep a 88ft doublet aloft?"

Table 2 on page 20-3 of the 19th edition of the "ARRL Antenna Book"
recommends 8 pounds tension for #18AWG hard-drawn copper wire.

Many issues are involved in tensioning. For calculations see Ed Laport`s
"Radio Antenna Engineering" page 346.

Best regards, Richard Harrison, KB5WZI

On Thu, 3 Mar 2005 11:40:09 -0600, (Richard
Harrison) wrote:
Table 2 on page 20-3 of the 19th edition of the "ARRL Antenna Book"
recommends 8 pounds tension for #18AWG hard-drawn copper wire.
Hi Richard,

From "Reference Data for Radio Engineers," the breaking load for #18
Hard-Drawn Copper is 85 pounds. The weight for 88 feet of this wire
(at 4.9 pounds per 1000') would be 0.43 pounds.

Many issues are involved in tensioning. For calculations see Ed Laport`s
"Radio Antenna Engineering" page 346.

Simple Trig would reveal that with an 88 foot catenary with a 5 foot
sag would develop roughly an angle of 6.5 degrees below the
horizontal. My earlier post employing arctan was in error. It is the
sine of the depressed angle used as a divider into half the weight (we
are presuming this is a symmetrical dipole) that describes the
tension. The sine of 6.5 degrees is 0.113. The tension in the wire
then becomes 0.215/0.113 pounds (1.9#).

Trying to flatten the catenary is where tension becomes increasingly
stressful. If you attempted to hold the catenary to within 1 foot of
flat, then this would be a depression of 1.3 degrees, resulting in a
divisor of 0.022. The tension in the wire then becomes 0.215/0.022
pounds (9.5#).

If we were to add in the weight of the transmission line drop, RG-58
comes in at roughly 0.029 pounds per foot or RG-11 as much as 0.096
pounds per foot.

Let's ballpark the last dipole with 1 foot sag to hold up 40 feet of
RG-11 (whatever the practicality, and irrespective of the center being
held up, we are considering only a two point suspension). The added
weight of the cable comes to 3.84 pounds. The tension in the wire
then becomes 2.135/0.022 pounds (97#).

The wire snaps.

73's
Richard Clark, KB7QHC

Richard Clark, KB7QHC wrote:
"Simple trig would reveal that with an 88 foot caternary with a 5 foot
sag would develop roughly an angle of 6;5 degrees below the horizontal."

I think we are about on the same page. Ed Laport says:
"If W is the equivalent total weight at the center, then the tension of
the triatic is:
T=W/2sinA

Where T and W are in identical units."

A is the deflection angle below an imaginary horizontal line.

Best regards, Richard Harrison, KB5WZI

Richard Clark wrote:
On Thu, 3 Mar 2005 11:40:09 -0600, (Richard
Harrison) wrote:
Table 2 on page 20-3 of the 19th edition of the "ARRL Antenna Book"
recommends 8 pounds tension for #18AWG hard-drawn copper wire.
Hi Richard,

From "Reference Data for Radio Engineers," the breaking load for #18
Hard-Drawn Copper is 85 pounds. The weight for 88 feet of this wire
(at 4.9 pounds per 1000') would be 0.43 pounds.

Many issues are involved in tensioning. For calculations see Ed
Laport`s
"Radio Antenna Engineering" page 346.

Simple Trig would reveal that with an 88 foot catenary with a 5 foot
sag would develop roughly an angle of 6.5 degrees below the
horizontal. My earlier post employing arctan was in error. It is
the
sine of the depressed angle used as a divider into half the weight
(we
are presuming this is a symmetrical dipole) that describes the
tension. The sine of 6.5 degrees is 0.113. The tension in the wire
then becomes 0.215/0.113 pounds (1.9#).

I think that if you do a really rigorous job of it you'll discover that
a hyperbolic function sneaks into the mix.


Trying to flatten the catenary is where tension becomes increasingly
stressful. If you attempted to hold the catenary to within 1 foot of
flat, then this would be a depression of 1.3 degrees, resulting in a
divisor of 0.022. The tension in the wire then becomes 0.215/0.022
pounds (9.5#).

If we were to add in the weight of the transmission line drop, RG-58
comes in at roughly 0.029 pounds per foot or RG-11 as much as 0.096
pounds per foot.

Let's ballpark the last dipole with 1 foot sag to hold up 40 feet of
RG-11 (whatever the practicality, and irrespective of the center
being
held up, we are considering only a two point suspension). The added
weight of the cable comes to 3.84 pounds. The tension in the wire
then becomes 2.135/0.022 pounds (97#).

The wire snaps.

73's
Richard Clark, KB7QHC

w3rv

On 3 Mar 2005 06:54:30 -0800, "Jim Miller" wrote:

What's the "typical" tension necessary to keep a 88ft doublet aloft?
I'll also be supporting the midpoint where the feedline attaches.
Insulated 18awg will be used as the antenna wire.

tnx
jtm

This isn't very scientific, but I just pull the wire tight until most
of the sag is gone. Never had one break. Also, if using trees for
support, a screen door spring at each end, between the tree and the
insulator, will keep the swaying tree from snapping the doublet.

Bob
k5qwg

"Bob Miller" wrote in message
...
On 3 Mar 2005 06:54:30 -0800, "Jim Miller" wrote:

What's the "typical" tension necessary to keep a 88ft doublet aloft?
I'll also be supporting the midpoint where the feedline attaches.
Insulated 18awg will be used as the antenna wire.

tnx
jtm

This isn't very scientific, but I just pull the wire tight until most
of the sag is gone. Never had one break. Also, if using trees for
support, a screen door spring at each end, between the tree and the
insulator, will keep the swaying tree from snapping the doublet.

Bob
k5qwg



I tried all that scientific formula stuff, and the KISS stuff. But neither
made the tree rats happy.

So my answer is; put it up until the squirels find the line and chew it in
half, then put it back up.

Dan/W4NTI

I remember reading somewhere about a cure for the squirrels. I believe it
was a mixture of cayenne pepper and molasses on the support ropes. I had a
cat once that snapped food out of your hand like a dog. Fed him a jalapeño
one day. Cured him.

"Dan/W4NTI" wrote in message
news:dgOVd.610
So my answer is; put it up until the squirels find the line and chew it
in
half, then put it back up.

Dan/W4NTI