My neighbor said the formula was 25 lbs to deflect the anti/drag wire 1/4". I tried it. It seemed like a lot of pressure on the wire. So I did some digging. I found a couple of things. The first was this formula:
T = (L * F) / 4 * X where
T is the tension on the wire.
L is the length of the wire.
F is the force in lbs applied at right angles at the mid point of the wire.
X is the deflection of the wire
There are some caveats. The diametre of the wire needs to be much smaller than the length so the stiffness of the wire is not a factor. The force is applied at right angles at the midpoint of the wire. And the force is small compared to the tension of the wire, ie: the deflection should be much much smaller than the wire length.
So plugging in what my neighbor gave me resulted in a tension of 1000 lbs = (40 * 25) / 4 * 0.25. It didn't make a lot of sense. I found this for a Stinson:
- Screenshot 2021-04-06 134748.jpg (178.16 KiB) Viewed 829 times
The tension was about a tenth of what I was told!! On some cub forums the number 1/2" deflection with 12-15 lbs of force was repeated several places. Dakota Cub was mentioned as the source. This comes out to about 240 - 300 lbs tension. My neighbor is building a J3. His wires are smaller than 3/16" I hope he doesn't have 1000 lbs tension on them.
So between the Stinson diagram and what Juerg Mueller said above makes a lot of sense.
PS: for those with a good ear, someone on the Internet said that the twang the wire makes should be "G" on a musical scale. I'm going to use my fish scale and 5 lbs force with 1/2" deflection which would put about 100 lbs tension of the wires. I also may check the twang. You never know. Build a Hatz is probably part artistry.
My neighbor said the formula was 25 lbs to deflect the anti/drag wire 1/4". I tried it. It seemed like a lot of pressure on the wire. So I did some digging. I found a couple of things. The first was this formula:
T = (L * F) / 4 * X where
T is the tension on the wire.
L is the length of the wire.
F is the force in lbs applied at right angles at the mid point of the wire.
X is the deflection of the wire
There are some caveats. The diametre of the wire needs to be much smaller than the length so the stiffness of the wire is not a factor. The force is applied at right angles at the midpoint of the wire. And the force is small compared to the tension of the wire, ie: the deflection should be much much smaller than the wire length.
So plugging in what my neighbor gave me resulted in a tension of 1000 lbs = (40 * 25) / 4 * 0.25. It didn't make a lot of sense. I found this for a Stinson:
[attachment=0]Screenshot 2021-04-06 134748.jpg[/attachment]
The tension was about a tenth of what I was told!! On some cub forums the number 1/2" deflection with 12-15 lbs of force was repeated several places. Dakota Cub was mentioned as the source. This comes out to about 240 - 300 lbs tension. My neighbor is building a J3. His wires are smaller than 3/16" I hope he doesn't have 1000 lbs tension on them.
So between the Stinson diagram and what Juerg Mueller said above makes a lot of sense.
PS: for those with a good ear, someone on the Internet said that the twang the wire makes should be "G" on a musical scale. I'm going to use my fish scale and 5 lbs force with 1/2" deflection which would put about 100 lbs tension of the wires. I also may check the twang. You never know. Build a Hatz is probably part artistry.