Lesson plan Grade 8 Social Studies & Civics
The Amberlock Canal: Freight Costs and Where Towns Grew
A grade 8 lesson in which students write and solve a freight-cost equation for a fictional canal, then use it to explain why canal towns grew faster.
Objectives
- Students will write a linear cost equation for canal freight from a stated toll and per-mile rate.
- Students will solve for the distance at which wagon freight and canal freight cost the same.
- Students will explain how a change in shipping cost affected where towns grew along a corridor.
Vocabulary
- Drainage divide
- A drainage divide is the high ground, like a ridge or hilltop, that separates one watershed from another. Rain that falls on opposite sides of a divide flows toward different streams.
Also: toll, ton-mile, summit level, freight
Teacher overview
Canals built in the 1820s and 1830s changed the cost of moving heavy goods over land, and the towns that sat on a canal often grew far faster than towns a few miles away. This lesson gives students a fictional canal with stated shipping rates so they can measure that advantage rather than assert it. The mathematical work is a linear equation with the variable on both sides, which students solve to find the trip length at which wagon freight and canal freight cost the same. A misconception worth naming before students begin is the belief that a lower per-mile rate always produces a lower total cost. Because the canal company charges a flat toll for every ton before the per-mile rate applies, short hauls stay cheaper by wagon, and students who ignore that constant term will reach the wrong break-even distance and the wrong conclusion about which towns benefited.
Procedure
Hook (5 min): Project the outline map of the fictional Amberlock Canal corridor, which runs 60 miles from Wexham to the port of Ledgeport. Ask students to predict how much a farmer saved by shipping a ton of grain on the canal instead of by wagon. Record three predictions on the board without comment.
Direct instruction (12 min): Explain that the canal crosses a drainage divide, so its highest section, called the summit level, is fed by a dam that holds back water and releases it into the canal. This is why canals could not simply follow a river. Then introduce the two rate structures students will work with: wagons charged a rate for every mile a ton traveled, while the canal company charged a smaller per-mile rate plus a flat toll for each ton entering the canal. Model the wagon cost for a 10-mile trip and the canal cost for the same trip using the handout's rates, and show that the canal is more expensive at that distance. Do not compute the 60-mile case; students will do that.
Guided practice (10 min): Working in pairs, students complete handout items 1 through 4 and compare answers. Circulate and check that pairs are multiplying the per-mile rate by the full 60 miles before adding the toll. Then poll the class with thumbs up or down on the question, "Does the canal cost less for every trip?" Expect a split. If fewer than half the class answers no, work through the 10-mile case again on the board before continuing.
Independent practice (15 min): Students complete handout items 5 through 7 on their own. Item 6 is the lesson's central task, and students should show the equation, not only the answer.
Closure and exit ticket (8 min): Students write their response to handout item 8 and turn in the full handout. Read a few responses aloud without names, and close by asking what a shipper in a town 12 miles off the canal would have paid to reach the canal in the first place.
Student Handout
The Amberlock Canal is a fictional waterway that opened in 1834. It runs 60 miles from the town of Wexham, on the Talloway River, to the port of Ledgeport. A dam on the Talloway feeds water into the canal's summit level, the highest stretch, where the canal crosses the divide between two watersheds.
Before the canal opened, shippers hauled goods to Ledgeport by wagon. A wagon trip cost $0.20 for each ton, for each mile traveled. On the canal, the boat company charged $0.03 for each ton, for each mile traveled, plus a flat toll of $4.00 per ton for entering the canal, no matter how far the ton traveled.
| Town | Location | Population in 1830 | Population in 1850 |
|---|---|---|---|
| Wexham | On the canal | 400 | 2,600 |
| Medlow | 12 miles from the canal | 450 | 700 |
- A farmer ships one ton of grain the full 60 miles by wagon. What does that trip cost?
- The same farmer ships one ton the full 60 miles by canal boat. What does that trip cost, including the toll?
- How much does the farmer save on one ton by shipping the full 60 miles on the canal instead of by wagon?
- A mill in Wexham ships 45 tons of flour to Ledgeport in one week. Using your answer to item 3, how much does the mill save on that shipment?
- Write an equation for the canal cost C, in dollars per ton, for a trip of m miles.
- Write an equation setting the wagon cost equal to the canal cost, then solve it to find the trip length at which the two cost the same. Round to the nearest tenth of a mile.
- How many more people lived in Wexham in 1850 than in 1830?
- Medlow sits 12 miles from the canal. Using your answer to item 3 and the wagon rate, explain why a shipper in Wexham gained more from the canal than a shipper in Medlow.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: $12.00.
The wagon rate applies to every mile: 0.20 x 60 = 12.00 dollars per ton. - Answer: $5.80.
The per-mile charge is 0.03 x 60 = 1.80, and the flat toll adds 4.00, so the trip costs 1.80 + 4.00 = 5.80 dollars per ton. - Answer: $6.20.
12.00 - 5.80 = 6.20 dollars saved on each ton. - Answer: $279.00.
45 tons x 6.20 dollars per ton = 279.00 dollars saved on the shipment. - Answer: C = 0.03m + 4.
The rate of change is 0.03 dollars per mile and the initial value is the 4.00 dollar toll, which is charged before the boat travels any distance. - Answer: about 23.5 miles.
0.20m = 0.03m + 4, so 0.17m = 4 and m = 23.53, which rounds to 23.5 miles. Below that distance the wagon costs less, because the toll has not yet been offset by the lower per-mile rate. - Answer: 2,200 people.
2,600 - 400 = 2,200 more people in 1850 than in 1830. - Answer: A Wexham shipper reached the canal without paying wagon rates first.
Goods leaving Wexham entered the canal directly and saved $6.20 on every ton. A Medlow shipper first paid 12 x 0.20 = 2.40 dollars per ton by wagon just to reach the canal, which ate into the savings. Accept any answer that uses the per-ton savings and the extra wagon cost.
Differentiation
Support
Give students a partly completed table for the fictional canal with the 10-mile and 30-mile costs already worked out under both rate structures, so item 6 becomes a matter of noticing where the two columns cross before solving algebraically. Allow calculators throughout, and provide the equation frame 0.20m = 0.03m + 4 with the first step shown.
Extension
Ask students to find the toll that would make the canal cheaper than the wagon for every trip longer than 10 miles in this fictional scenario, and to explain what such a toll would have meant for the canal company's revenue.
ELL supports
Pre-teach "toll," "per ton," "summit level," and "drainage divide" with a labeled cross-section drawing of a canal crossing a ridge. Provide sentence frames for item 8: "A shipper in Wexham saved ___ on each ton because ___. A shipper in Medlow saved less because ___."
Extensions
Have students research how a real canal in their own state or region was financed and who paid for it, then write a short comparison with the fictional Amberlock Canal's toll system. A second option is to map the towns within 10 miles of that real canal and compare their growth between two census years using published population figures.