**Fractal branching ultra-dexterous robots (Bush robots)** #

NASA ACRP Quarterly Report, February 1998, Hans Moravec

**NASA****
ADVANCED CONCEPTS RESEARCH PROJECTS
**PR-Number 10-86888 Appropriation: 806/70110

CMU Cooperative Agreement NCC7-7

Quarterly Report

for

December 1, 1997 - February 27, 1998

(Bush robots)

Jesse Easudes

Robotics Institute

5000 Forbes Avenue

Pittsburgh, PA 15213

USA

February 20, 1998

**Table of Contents**

**Summary** 3

**Section 1:** Joint Model 4

**Section 2:** Control Wire Stresses 5

**Section 3:** General Excursion to Force Relation 8

**Section 4:** Test Joint Mechanics 9

**Section 5:** Test Joint Electrical and Thermal Design 12

**Section 6:** Observations 13

Summary

This is the fifth quarterly report for “Fractal branching ultra-dexterous robots.”

In this quarter we designed and built a prototype segment for a maximally simple shape-memory alloy actuated bush robot. Our experiences with this model are being used to guide a more competent design that will be scaled to multiple levels.

An original branch of our preferred (

**L**_{upper} =
**L** ( 3^{–1/2} + 3^{–2/2}
+ 3^{–3/2} + ... + 3^{–i/2}
+ ... ) =

**L** / (sqrt(3) – 1) = 1.366 **L**

At the i’th level, each branch has cross-sectional area
proportional to 3^{–i}, but there are 3^{i}
branches with that cross section, so the net area, given by the
product of those two numbers, remains constant. Thus, with perfect
scaling, the mass of tree carried above any given branch is no more
than 1.366 times the mass of the branch itself.

For purposes of developing the a working joint, we can therefore
simulate the loading of a complete bush by extending the branch an
additional 1.366 times its basic length, also ensuring that the
extension masses 1.366 times as much. This is a very modest
requirement for mechanical design of a joint, smaller even than the
safety margins of two or more that would normally be
incorporated.

Our prototype single joint will thus probably appear as
follows:

The diameter of nitinol wire will be chosen to be able to lift, when
contracting to its natural length, at least twice the weight of the
branch plus ballast. We will be using thin aluminum sections for the
branches. Our largest branch may be about 10 cm in length, and should
weigh only a fraction of a kilogram. Wire with diameter of 0.25 mm,
which can lift about one kilogram, should be more than adequate. This
same wire requires a restoring force of about 0.2 kg to stretch it
when cold.

Higher, smaller, levels of the bush will use correspondingly thinner
wire, the cross section reduced by a factor of 1/3 for each level, the
diameter by 1/sqrt(3).

**2: Control Wire Stresses**

We analyse the loading on the control wires due to weight of the bush
structure by considering a branch extended horizontally. In the
following diagram, **L** is the distance from the branch base to
the attachment point of the control wires, **B** is the distance
from the branch pivot of the base of the wires, **C** is the
distance of the folded bush center of gravity from the pivot.
**W** is the weight of the full bush above the pivot. The tilt
angle of the segment below the horizontal is designated by
.

Neglecting the lower wire, note that the upper wire must exert a
torque about the pivot cancelling the torque induced by the weight
**W**. The torque caused by the weight has magnitude **W**
**C** cos . If the tension in the wire is
designated by **T**, the wire exerts torque

In modeling the folded upper layers of the bush, we had concluded
that the subtree above the length **L** segment was equivalent to
extending the segment by an amount 1.366 **L**. This makes the
length **C** = 1.183 **L**. We had also noted in the last
report that letting **B** = **L**/20 allows a +/– 30 degree
deflection with 5% wire elongation. Choose this value for **B**.
Then the torque equivalence becomes:

From which we find the ratio of **T** to **W**:

This produces the following plot of wire tension as a function of
, showing a maximum force **T** of under 25
times **W**, even when the angle is brought beyond the 30°
limit, to 90° vertical:

The previous analysis orients the base horizontally, which causes the
gravity torque to fall off as the branch deviates from its central
orientation. A worse case occurs if the base is rotated to keep the
branch horizontal at all times:

In this case,

and the curve looks as follows, with T still staying below 30 W
when the deflection is within +/– 30° :

A factor still missing in the above analyses is the force needed to
stretch the opposing wire as the supporting “power” wire
contracts. A typical nitinol wire when cold can be stretched with
about 1/6 (i.e. less than 17%) the force it exerts when hot. This is
small enough to be absorbable in a safety factor of two added to the
above analysis. Thus we plan to use nitinol wire whose contraction
strength is 60 times the weight of the bushlet it moves. The cold
strength of a stretched wire would, in that case, not be enough to
support the weight of its bush, if the forces in the above analysis
were reversed. In that case, a mechanical stop at 30 degree deviation
at the pivot of the branch could provide the necessary
support.

**3: General Excursion to Force Relation**

The previous section examined the stresses for a particular branch
excursion, +/– 30 degrees, for which the necessary wire
contraction force was equal to 30 times the subbush weight (or 60
times, with a 2x safety factor).

To see how the wire stress depends on the desired excursion, we note
from the special analysis that the maximum stress occurs at the point
of maximum stretch of the wire (i.e. the maximum value of angle ). If we choose length units to make the branch
length **L** = 1, last quarter’s report gives the maximum
excursion as:

where **ER** is (**E**^{2} – 1)/(
**E**^{2} + 1 ), and **E** is the length ratio of
expanded to contracted wires. Our nitinol documentation suggests
**E** = 1.05 allows long-term operation.

We can rewrite the relation from the previous section to reveal the
general form for the wire force as a function of angle:

where **C** = (1 + 1/(sqrt(3)-1))/2 = 1.183 is the relative
position of the folded subbush center of gravity.

Substituting * _{max}* for in the
expression for

where **ER** is the ratio (**E**^{2} –
1)/(**E**^{2} + 1). This results in the following
relationship between maximum wire force **T / W** and maximum
angular excursion
* _{max}*.

**4: Test Joint Mechanics**

We’ve constructed a one-axis test bush segment joint out of
threaded metal spacers and a hinge. Its basic length is 9 cm. With
additional spacers to simulate the upper part of the bush, its length
grows to 22 cm. The combined weight of this assembly is 115 grams.
We mounted two control wires, each with a lateral offset of 0.5 cm
from the side of the branch, to provide about +/– 30° of
deflection with 5% wire elongation. This setup requires a contraction
strength of about 7,000 grams, to support the full weight laterally,
with a safety factor of two. A typical nitinol alloy (for instance
the one used in Flexinol “Muscle Wire”) exerts this force in
a wire about 0.66 mm in diameter.

We have chosen low-temperature Nitinol alloys with a 70° C to 90° C transition temperatures, compared to typical ambient temperatures of 20° C to 30° C. Low temperature and thickness in wires both act to increase the cooling time. The table below give cooling time in still air. The time can be reduced tenfold by fast ventilation.

Here is a table of approximate heating currents and cooling times:

Wire Resistance Heating Contraction Cooling

Diameter (ohm/meter) Current Force Time

.025 mm 1770 20 ma 7 g 1.1 sec

.050 mm 510 50 ma 35 g 1.3 sec

.100 mm 150 180 ma 150 g 1.8 sec

.150 mm 50 400 ma 330 g 3.0 sec

.250 mm 20 1 A 930 g 6.7 sec

.350 mm 9.0 2 A 2 kg 11 sec

.500 mm 5.1 4 A 4 kg 19 sec

.660 mm 2.54 7 A 7 kg 30 sec

.750 mm 2.2 8 A 9 kg 36 sec

1.00 mm 1.5 15 A 15 kg 55 sec

The long cooling time of our 0.66 mm test joint wire may encourage us to add an air-cooling fan, to reduce the time to a few seconds. It can be heated in a few milliseconds by a current of a few amps. A 10 cm length has a resistance of only about 0.25 ohms.

We plan to use the driving circuit pictured below. A 5 volt supply will suffice to drive many amps into the 0.25 ohms. To provide seven amperes for a millisecond, the capacitor C must store at least seven millicoulombs. At an average of two volts during discharge, this would require about 3,500 microfarads. We are using a 5 volt, 3,500 microfarad capacitor to power our test branch.

The charging resistor should “refill” the capacitor in the
30 second cooling time of the wire. The RC time constant should thus
be perhaps 10 seconds. Since C is 3,500 microfarads, R is chosen to
be 3 Kohms.

**6: Observations**

The Nitinol wire is hard and resists deformation. We found it
difficult to anchor: it tended to slip out of regular pressure clamps.
We were successful with one technique, inserting the wire through a
small hole drilled through the side of a threaded spacer. The
slightly bent nitinol wire can be securely pinched by a screw in the
spacer:

The wire also resists bends, so it was necessary to pivot the anchors
so the released the wire very close to its intended path. Even so,
the wire exhibits bowing, even under stress. In future we will fasten
pivoting eyelets to both ends of each wire. It will then be much
easier to make straight runs of wire of exactly the correct length.
These eyelets will be held in place by a pivot pin, forming a
hinge: